Solve Using the Quadratic Formula 4x^2-2x-5=0

$4 x^{2} - 2 x - 5 = 0$

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 $a = 4$ , $b = - 2$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16 \cdot - 5}}{2 \cdot 4}$

Multiply $- 16$ by $- 5$ .

$x = \frac{2 \pm \sqrt{4 + 80}}{2 \cdot 4}$

Add $4$ and $80$ .

$x = \frac{2 \pm \sqrt{84}}{2 \cdot 4}$

Rewrite $84$ as $2^{2} \cdot 21$ .

Factor $4$ out of $84$ .

$x = \frac{2 \pm \sqrt{4 (21)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 21}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{21}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{2 \pm 2 \sqrt{21}}{8}$

Simplify $\frac{2 \pm 2 \sqrt{21}}{8}$ .

$x = \frac{1 \pm \sqrt{21}}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{21}}{4}$

Decimal Form:

$x = 1.39564392 \dots, - 0.89564392 \dots$