Solve for q 6q^2-13q+6=2q^2

$6 q^{2} - 13 q + 6 = 2 q^{2}$

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

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

$6 q^{2} - 13 q + 6 - 2 q^{2} = 0$

Subtract $2 q^{2}$ from $6 q^{2}$ .

$4 q^{2} - 13 q + 6 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 13$ , and $c = 6$ into the quadratic formula and solve for $q$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (4 \cdot 6)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$q = \frac{13 \pm \sqrt{169 - 4 \cdot 4 \cdot 6}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$q = \frac{13 \pm \sqrt{169 - 16 \cdot 6}}{2 \cdot 4}$

Multiply $- 16$ by $6$ .

$q = \frac{13 \pm \sqrt{169 - 96}}{2 \cdot 4}$

Subtract $96$ from $169$ .

$q = \frac{13 \pm \sqrt{73}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$q = \frac{13 \pm \sqrt{73}}{8}$

The final answer is the combination of both solutions.

$q = \frac{13 + \sqrt{73}}{8}, \frac{13 - \sqrt{73}}{8}$

The result can be shown in multiple forms.

Exact Form:

$q = \frac{13 + \sqrt{73}}{8}, \frac{13 - \sqrt{73}}{8}$

Decimal Form:

$q = 2.69300046 \dots, 0.55699953 \dots$