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

$f (x) = 3 x^{2} - 3 x + 5$

Step 1

Set $3 x^{2} - 3 x + 5$ equal to $0$ .

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

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 5$ .

Multiply $- 4$ by $3$ .

$x = \frac{3 \pm \sqrt{9 - 12 \cdot 5}}{2 \cdot 3}$

Multiply $- 12$ by $5$ .

$x = \frac{3 \pm \sqrt{9 - 60}}{2 \cdot 3}$

Subtract $60$ from $9$ .

$x = \frac{3 \pm \sqrt{- 51}}{2 \cdot 3}$

Rewrite $- 51$ as $- 1 (51)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 51}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (51)}$ as $\sqrt{- 1} \cdot \sqrt{51}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{51}}{2 \cdot 3}$

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

$x = \frac{3 \pm i \sqrt{51}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{3 \pm i \sqrt{51}}{6}$

The final answer is the combination of both solutions.

$x = \frac{3 + i \sqrt{51}}{6}, \frac{3 - i \sqrt{51}}{6}$

$x = \frac{3 \pm i \sqrt{51}}{6}$

Step 3