Solve Using the Quadratic Formula 5x^2=2x+4

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

Step 1

Move all the expressions to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

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

Subtract $4$ from both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $5$ .

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

Multiply $- 20$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 80}}{2 \cdot 5}$

Add $4$ and $80$ .

$x = \frac{2 \pm \sqrt{84}}{2 \cdot 5}$

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

Factor $4$ out of $84$ .

$x = \frac{2 \pm \sqrt{4 (21)}}{2 \cdot 5}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 21}}{2 \cdot 5}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{21}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{2 \pm 2 \sqrt{21}}{10}$

Simplify $\frac{2 \pm 2 \sqrt{21}}{10}$ .

$x = \frac{1 \pm \sqrt{21}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{21}}{5}$

Decimal Form:

$x = 1.11651513 \dots, - 0.71651513 \dots$