Solve Using the Quadratic Formula 5x^2-36x+7=0

$5 x^{2} - 36 x + 7 = 0$

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 $a = 5$ , $b = - 36$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{36 \pm \sqrt{{(- 36)}^{2} - 4 \cdot (5 \cdot 7)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{36 \pm \sqrt{1296 - 4 \cdot 5 \cdot 7}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 7$ .

Multiply $- 4$ by $5$ .

$x = \frac{36 \pm \sqrt{1296 - 20 \cdot 7}}{2 \cdot 5}$

Multiply $- 20$ by $7$ .

$x = \frac{36 \pm \sqrt{1296 - 140}}{2 \cdot 5}$

Subtract $140$ from $1296$ .

$x = \frac{36 \pm \sqrt{1156}}{2 \cdot 5}$

Rewrite $1156$ as $34^{2}$ .

$x = \frac{36 \pm \sqrt{34^{2}}}{2 \cdot 5}$

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

$x = \frac{36 \pm 34}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{36 \pm 34}{10}$

Simplify $\frac{36 \pm 34}{10}$ .

$x = \frac{18 \pm 17}{5}$

Step 4

The final answer is the combination of both solutions.

$x = 7, \frac{1}{5}$