Find the Roots (Zeros) 4x^2-36x=-73

$4 x^{2} - 36 x = - 73$

Add $73$ to both sides of the equation.

$4 x^{2} - 36 x + 73 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 36$ , and $c = 73$ into the quadratic formula and solve for $x$ .

$\frac{36 \pm \sqrt{{(- 36)}^{2} - 4 \cdot (4 \cdot 73)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 36$ to the power of $2$ .

$x = \frac{36 \pm \sqrt{1296 - 4 \cdot 4 \cdot 73}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{36 \pm \sqrt{1296 - 16 \cdot 73}}{2 \cdot 4}$

Multiply $- 16$ by $73$ .

$x = \frac{36 \pm \sqrt{1296 - 1168}}{2 \cdot 4}$

Subtract $1168$ from $1296$ .

$x = \frac{36 \pm \sqrt{128}}{2 \cdot 4}$

Rewrite $128$ as $8^{2} \cdot 2$ .

Factor $64$ out of $128$ .

$x = \frac{36 \pm \sqrt{64 (2)}}{2 \cdot 4}$

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

$x = \frac{36 \pm \sqrt{8^{2} \cdot 2}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{36 \pm 8 \sqrt{2}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{36 \pm 8 \sqrt{2}}{8}$

Simplify $\frac{36 \pm 8 \sqrt{2}}{8}$ .

$x = \frac{9 \pm 2 \sqrt{2}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{9 \pm 2 \sqrt{2}}{2}$

Decimal Form:

$x = 5.91421356 \dots, 3.08578643 \dots$