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

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $7$ .

$x = \frac{3 \pm \sqrt{9 - 28 \cdot - 5}}{2 \cdot 7}$

Multiply $- 28$ by $- 5$ .

$x = \frac{3 \pm \sqrt{9 + 140}}{2 \cdot 7}$

Add $9$ and $140$ .

$x = \frac{3 \pm \sqrt{149}}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{3 \pm \sqrt{149}}{14}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{149}}{14}$

Decimal Form:

$x = 1.08618254 \dots, - 0.65761111 \dots$