Algebra Examples

$f (x) = x^{2} - 2 x + 3$ , $f (x) = - 2 x + 19$

Step 1

Substitute $- 2 x + 19$ for $f (x)$ .

$- 2 x + 19 = x^{2} - 2 x + 3$

Step 2

Solve for $x$ .

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Step 2.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 2 x + 3 = - 2 x + 19$

Step 2.2

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

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Step 2.2.1

Add $2 x$ to both sides of the equation.

$x^{2} - 2 x + 3 + 2 x = 19$

Step 2.2.2

Combine the opposite terms in $x^{2} - 2 x + 3 + 2 x$ .

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Step 2.2.2.1

Add $- 2 x$ and $2 x$ .

$x^{2} + 0 + 3 = 19$

Step 2.2.2.2

Add $x^{2}$ and $0$ .

$x^{2} + 3 = 19$

Step 2.3

Move all terms not containing $x$ to the right side of the equation.

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Step 2.3.1

Subtract $3$ from both sides of the equation.

$x^{2} = 19 - 3$

Step 2.3.2

Subtract $3$ from $19$ .

$x^{2} = 16$

Step 2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Step 2.5

Simplify $\pm \sqrt{16}$ .

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Step 2.5.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Step 2.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

Step 2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Step 2.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

Step 2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$