Find the Roots (Zeros) 15x^2-5x+6=9x^2

$15 x^{2} - 5 x + 6 = 9 x^{2}$

Step 1

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

Subtract $9 x^{2}$ from both sides of the equation.

$15 x^{2} - 5 x + 6 - 9 x^{2} = 0$

Subtract $9 x^{2}$ from $15 x^{2}$ .

$6 x^{2} - 5 x + 6 = 0$

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 $a = 6$ , $b = - 5$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (6 \cdot 6)}}{2 \cdot 6}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 6 \cdot 6}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 6$ .

Multiply $- 4$ by $6$ .

$x = \frac{5 \pm \sqrt{25 - 24 \cdot 6}}{2 \cdot 6}$

Multiply $- 24$ by $6$ .

$x = \frac{5 \pm \sqrt{25 - 144}}{2 \cdot 6}$

Subtract $144$ from $25$ .

$x = \frac{5 \pm \sqrt{- 119}}{2 \cdot 6}$

Rewrite $- 119$ as $- 1 (119)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 119}}{2 \cdot 6}$

Rewrite $\sqrt{- 1 (119)}$ as $\sqrt{- 1} \cdot \sqrt{119}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{119}}{2 \cdot 6}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{5 \pm i \sqrt{119}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{5 \pm i \sqrt{119}}{12}$

Step 5