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

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

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 $a = 5$ , $b = - 2$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $5$ .

$x = \frac{2 \pm \sqrt{4 - 20 \cdot - 3}}{2 \cdot 5}$

Multiply $- 20$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 60}}{2 \cdot 5}$

Add $4$ and $60$ .

$x = \frac{2 \pm \sqrt{64}}{2 \cdot 5}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{2 \pm \sqrt{8^{2}}}{2 \cdot 5}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm 8}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{2 \pm 8}{10}$

Simplify $\frac{2 \pm 8}{10}$ .

$x = \frac{1 \pm 4}{5}$

Step 4

The final answer is the combination of both solutions.

$x = 1, - \frac{3}{5}$