Solve Using the Quadratic Formula 5x^2+21x-20=0

$5 x^{2} + 21 x - 20 = 0$

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 $a = 5$ , $b = 21$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{- 21 \pm \sqrt{21^{2} - 4 \cdot (5 \cdot - 20)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $21$ to the power of $2$ .

$x = \frac{- 21 \pm \sqrt{441 - 4 \cdot 5 \cdot - 20}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 20$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 21 \pm \sqrt{441 - 20 \cdot - 20}}{2 \cdot 5}$

Multiply $- 20$ by $- 20$ .

$x = \frac{- 21 \pm \sqrt{441 + 400}}{2 \cdot 5}$

Add $441$ and $400$ .

$x = \frac{- 21 \pm \sqrt{841}}{2 \cdot 5}$

Rewrite $841$ as $29^{2}$ .

$x = \frac{- 21 \pm \sqrt{29^{2}}}{2 \cdot 5}$

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

$x = \frac{- 21 \pm 29}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 21 \pm 29}{10}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{4}{5}, - 5$