Solve Using the Quadratic Formula 5x^2-3x+7=0

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 5 \cdot 7}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 7$ .

Multiply $- 4$ by $5$ .

$x = \frac{3 \pm \sqrt{9 - 20 \cdot 7}}{2 \cdot 5}$

Multiply $- 20$ by $7$ .

$x = \frac{3 \pm \sqrt{9 - 140}}{2 \cdot 5}$

Subtract $140$ from $9$ .

$x = \frac{3 \pm \sqrt{- 131}}{2 \cdot 5}$

Rewrite $- 131$ as $- 1 (131)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 131}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (131)}$ as $\sqrt{- 1} \cdot \sqrt{131}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{131}}{2 \cdot 5}$

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

$x = \frac{3 \pm i \sqrt{131}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{3 \pm i \sqrt{131}}{10}$