Find the Roots (Zeros) 4x^2=8x-7

$4 x^{2} = 8 x - 7$

Subtract $8 x$ from both sides of the equation.

$4 x^{2} - 8 x = - 7$

Add $7$ to both sides of the equation.

$4 x^{2} - 8 x + 7 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 8$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (4 \cdot 7)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 4 \cdot 7}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{8 \pm \sqrt{64 - 16 \cdot 7}}{2 \cdot 4}$

Multiply $- 16$ by $7$ .

$x = \frac{8 \pm \sqrt{64 - 112}}{2 \cdot 4}$

Subtract $112$ from $64$ .

$x = \frac{8 \pm \sqrt{- 48}}{2 \cdot 4}$

Rewrite $- 48$ as $- 1 (48)$ .

$x = \frac{8 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$x = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{8 \pm i \cdot \sqrt{48}}{2 \cdot 4}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \frac{8 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 4}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{8 \pm i \cdot \sqrt{4^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{8 \pm i \cdot (4 \sqrt{3})}{2 \cdot 4}$

Move $4$ to the left of $i$ .

$x = \frac{8 \pm 4 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{8 \pm 4 i \sqrt{3}}{8}$

Simplify $\frac{8 \pm 4 i \sqrt{3}}{8}$ .

$x = \frac{2 \pm i \sqrt{3}}{2}$