Find the Roots (Zeros) 4x^2-28x=-53

$4 x^{2} - 28 x = - 53$

Add $53$ to both sides of the equation.

$4 x^{2} - 28 x + 53 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 28$ , and $c = 53$ into the quadratic formula and solve for $x$ .

$\frac{28 \pm \sqrt{{(- 28)}^{2} - 4 \cdot (4 \cdot 53)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 28$ to the power of $2$ .

$x = \frac{28 \pm \sqrt{784 - 4 \cdot 4 \cdot 53}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{28 \pm \sqrt{784 - 16 \cdot 53}}{2 \cdot 4}$

Multiply $- 16$ by $53$ .

$x = \frac{28 \pm \sqrt{784 - 848}}{2 \cdot 4}$

Subtract $848$ from $784$ .

$x = \frac{28 \pm \sqrt{- 64}}{2 \cdot 4}$

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{28 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{28 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 4}$

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

$x = \frac{28 \pm i \cdot \sqrt{64}}{2 \cdot 4}$

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

$x = \frac{28 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 4}$

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

$x = \frac{28 \pm i \cdot 8}{2 \cdot 4}$

Move $8$ to the left of $i$ .

$x = \frac{28 \pm 8 i}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{28 \pm 8 i}{8}$

Simplify $\frac{28 \pm 8 i}{8}$ .

$x = \frac{7 \pm 2 i}{2}$