Solve Using the Quadratic Formula 5x^2-35x+60=0

$5 x^{2} - 35 x + 60 = 0$

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 $a = 5$ , $b = - 35$ , and $c = 60$ into the quadratic formula and solve for $x$ .

$\frac{35 \pm \sqrt{{(- 35)}^{2} - 4 \cdot (5 \cdot 60)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 35$ to the power of $2$ .

$x = \frac{35 \pm \sqrt{1225 - 4 \cdot 5 \cdot 60}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 60$ .

Multiply $- 4$ by $5$ .

$x = \frac{35 \pm \sqrt{1225 - 20 \cdot 60}}{2 \cdot 5}$

Multiply $- 20$ by $60$ .

$x = \frac{35 \pm \sqrt{1225 - 1200}}{2 \cdot 5}$

Subtract $1200$ from $1225$ .

$x = \frac{35 \pm \sqrt{25}}{2 \cdot 5}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{35 \pm \sqrt{5^{2}}}{2 \cdot 5}$

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

$x = \frac{35 \pm 5}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{35 \pm 5}{10}$

Simplify $\frac{35 \pm 5}{10}$ .

$x = \frac{7 \pm 1}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 4, 3$