Algebra Examples

$4 x^{2} + 6 x + 35 = 16 x^{2} + 7 x$

Step 1

Move all terms containing $x$ to the left side of the equation.

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Step 1.1

Subtract $16 x^{2}$ from both sides of the equation.

$4 x^{2} + 6 x + 35 - 16 x^{2} = 7 x$

Step 1.2

Subtract $7 x$ from both sides of the equation.

$4 x^{2} + 6 x + 35 - 16 x^{2} - 7 x = 0$

Step 1.3

Subtract $16 x^{2}$ from $4 x^{2}$ .

$- 12 x^{2} + 6 x + 35 - 7 x = 0$

Step 1.4

Subtract $7 x$ from $6 x$ .

$- 12 x^{2} - x + 35 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $- 1$ out of $- 12 x^{2} - x + 35$ .

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Step 2.1.1

Factor $- 1$ out of $- 12 x^{2}$ .

$- (12 x^{2}) - x + 35 = 0$

Step 2.1.2

Factor $- 1$ out of $- x$ .

$- (12 x^{2}) - (x) + 35 = 0$

Step 2.1.3

Rewrite $35$ as $- 1 (- 35)$ .

$- (12 x^{2}) - (x) - 1 \cdot - 35 = 0$

Step 2.1.4

Factor $- 1$ out of $- (12 x^{2}) - (x)$ .

$- (12 x^{2} + x) - 1 \cdot - 35 = 0$

Step 2.1.5

Factor $- 1$ out of $- (12 x^{2} + x) - 1 (- 35)$ .

$- (12 x^{2} + x - 35) = 0$

Step 2.2

Factor.

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Step 2.2.1

Factor by grouping.

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Step 2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 12 \cdot - 35 = - 420$ and whose sum is $b = 1$ .

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Step 2.2.1.1.1

Multiply by $1$ .

$- (12 x^{2} + 1 x - 35) = 0$

Step 2.2.1.1.2

Rewrite $1$ as $- 20$ plus $21$

$- (12 x^{2} + (- 20 + 21) x - 35) = 0$

Step 2.2.1.1.3

Apply the distributive property.

$- (12 x^{2} - 20 x + 21 x - 35) = 0$

Step 2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.1.2.1

Group the first two terms and the last two terms.

$- ((12 x^{2} - 20 x) + 21 x - 35) = 0$

Step 2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$- (4 x (3 x - 5) + 7 (3 x - 5)) = 0$

Step 2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 5$ .

$- ((3 x - 5) (4 x + 7)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$- (3 x - 5) (4 x + 7) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 5 = 0$

$4 x + 7 = 0$

Step 4

Set $3 x - 5$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Step 4.2

Solve $3 x - 5 = 0$ for $x$ .

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Step 4.2.1

Add $5$ to both sides of the equation.

$3 x = 5$

Step 4.2.2

Divide each term in $3 x = 5$ by $3$ and simplify.

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Step 4.2.2.1

Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $3$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Step 4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{3}$

Step 5

Set $4 x + 7$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $4 x + 7$ equal to $0$ .

$4 x + 7 = 0$

Step 5.2

Solve $4 x + 7 = 0$ for $x$ .

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Step 5.2.1

Subtract $7$ from both sides of the equation.

$4 x = - 7$

Step 5.2.2

Divide each term in $4 x = - 7$ by $4$ and simplify.

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Step 5.2.2.1

Divide each term in $4 x = - 7$ by $4$ .

$\frac{4 x}{4} = \frac{- 7}{4}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 7}{4}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 7}{4}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{7}{4}$

Step 6

The final solution is all the values that make $- (3 x - 5) (4 x + 7) = 0$ true.

$x = \frac{5}{3}, - \frac{7}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{3}, - \frac{7}{4}$

Decimal Form:

$x = 1.\overline{6}, - 1.75$

Mixed Number Form:

$x = 1 \frac{2}{3}, - 1 \frac{3}{4}$