Find the Roots (Zeros) 3x^2=4x-15

$3 x^{2} = 4 x - 15$

Subtract $4 x$ from both sides of the equation.

$3 x^{2} - 4 x = - 15$

Add $15$ to both sides of the equation.

$3 x^{2} - 4 x + 15 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 4$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (3 \cdot 15)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 3 \cdot 15}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$x = \frac{4 \pm \sqrt{16 - 12 \cdot 15}}{2 \cdot 3}$

Multiply $- 12$ by $15$ .

$x = \frac{4 \pm \sqrt{16 - 180}}{2 \cdot 3}$

Subtract $180$ from $16$ .

$x = \frac{4 \pm \sqrt{- 164}}{2 \cdot 3}$

Rewrite $- 164$ as $- 1 (164)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 164}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (164)}$ as $\sqrt{- 1} \cdot \sqrt{164}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{164}}{2 \cdot 3}$

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

$x = \frac{4 \pm i \cdot \sqrt{164}}{2 \cdot 3}$

Rewrite $164$ as $2^{2} \cdot 41$ .

Factor $4$ out of $164$ .

$x = \frac{4 \pm i \cdot \sqrt{4 (41)}}{2 \cdot 3}$

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 41}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (2 \sqrt{41})}{2 \cdot 3}$

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i \sqrt{41}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{4 \pm 2 i \sqrt{41}}{6}$

Simplify $\frac{4 \pm 2 i \sqrt{41}}{6}$ .

$x = \frac{2 \pm i \sqrt{41}}{3}$