Find the Roots (Zeros) 2x^2+4x-5=0

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

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

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

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

Multiply $- 4 \cdot 2 \cdot - 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{- 4 \pm \sqrt{16 + 40}}{2 \cdot 2}$

Add $16$ and $40$ .

$x = \frac{- 4 \pm \sqrt{56}}{2 \cdot 2}$

Rewrite $56$ as $2^{2} \cdot 14$ .

Factor $4$ out of $56$ .

$x = \frac{- 4 \pm \sqrt{4 (14)}}{2 \cdot 2}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 14}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{14}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 4 \pm 2 \sqrt{14}}{4}$

Simplify $\frac{- 4 \pm 2 \sqrt{14}}{4}$ .

$x = \frac{- 2 \pm \sqrt{14}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 2 \pm \sqrt{14}}{2}$

Decimal Form:

$x = 0.87082869 \dots, - 2.87082869 \dots$

Step 5