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

$f (x) = 2 x^{2} - 3 x + 7$

Step 1

Set $2 x^{2} - 3 x + 7$ equal to $0$ .

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

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 $a = 2$ , $b = - 3$ , and $c = 7$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 7}}{2 \cdot 2}$

Multiply $- 8$ by $7$ .

$x = \frac{3 \pm \sqrt{9 - 56}}{2 \cdot 2}$

Subtract $56$ from $9$ .

$x = \frac{3 \pm \sqrt{- 47}}{2 \cdot 2}$

Rewrite $- 47$ as $- 1 (47)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 47}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (47)}$ as $\sqrt{- 1} \cdot \sqrt{47}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{47}}{2 \cdot 2}$

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

$x = \frac{3 \pm i \sqrt{47}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm i \sqrt{47}}{4}$

The final answer is the combination of both solutions.

$x = \frac{3 + i \sqrt{47}}{4}, \frac{3 - i \sqrt{47}}{4}$

$x = \frac{3 \pm i \sqrt{47}}{4}$

Step 3