Find the Roots (Zeros) 3x^2-7x+5=0

$3 x^{2} - 7 x + 5 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 7$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (3 \cdot 5)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

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

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$x = \frac{7 \pm \sqrt{49 - 12 \cdot 5}}{2 \cdot 3}$

Multiply $- 12$ by $5$ .

$x = \frac{7 \pm \sqrt{49 - 60}}{2 \cdot 3}$

Subtract $60$ from $49$ .

$x = \frac{7 \pm \sqrt{- 11}}{2 \cdot 3}$

Rewrite $- 11$ as $- 1 (11)$ .

$x = \frac{7 \pm \sqrt{- 1 \cdot 11}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (11)}$ as $\sqrt{- 1} \cdot \sqrt{11}$ .

$x = \frac{7 \pm \sqrt{- 1} \cdot \sqrt{11}}{2 \cdot 3}$

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

$x = \frac{7 \pm i \sqrt{11}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{7 \pm i \sqrt{11}}{6}$