Algebra Examples

$- 2 v^{2} - v + 12 = - 3 v^{2} + 6 v$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Move all the expressions to the left side of the equation.

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Step 1.1.1

Add $3 v^{2}$ to both sides of the equation.

$- 2 v^{2} - v + 12 + 3 v^{2} = 6 v$

Step 1.1.2

Subtract $6 v$ from both sides of the equation.

$- 2 v^{2} - v + 12 + 3 v^{2} - 6 v = 0$

Step 1.2

Simplify $- 2 v^{2} - v + 12 + 3 v^{2} - 6 v$ .

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Step 1.2.1

Add $- 2 v^{2}$ and $3 v^{2}$ .

$v^{2} - v + 12 - 6 v = 0$

Step 1.2.2

Subtract $6 v$ from $- v$ .

$v^{2} - 7 v + 12 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 7$ , and $c = 12$ into the quadratic formula and solve for $v$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot 12)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 7$ to the power of $2$ .

$v = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 12$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$v = \frac{7 \pm \sqrt{49 - 4 \cdot 12}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $12$ .

$v = \frac{7 \pm \sqrt{49 - 48}}{2 \cdot 1}$

Step 4.1.3

Subtract $48$ from $49$ .

$v = \frac{7 \pm \sqrt{1}}{2 \cdot 1}$

Step 4.1.4

Any root of $1$ is $1$ .

$v = \frac{7 \pm 1}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$v = \frac{7 \pm 1}{2}$

Step 5

The final answer is the combination of both solutions.

$v = 4, 3$