Solve Using the Quadratic Formula 3x^2+11x=20

$3 x^{2} + 11 x = 20$

Step 1

Subtract $20$ from both sides of the equation.

$3 x^{2} + 11 x - 20 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 3$ , $b = 11$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (3 \cdot - 20)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot 3 \cdot - 20}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 20$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 11 \pm \sqrt{121 - 12 \cdot - 20}}{2 \cdot 3}$

Multiply $- 12$ by $- 20$ .

$x = \frac{- 11 \pm \sqrt{121 + 240}}{2 \cdot 3}$

Add $121$ and $240$ .

$x = \frac{- 11 \pm \sqrt{361}}{2 \cdot 3}$

Rewrite $361$ as $19^{2}$ .

$x = \frac{- 11 \pm \sqrt{19^{2}}}{2 \cdot 3}$

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

$x = \frac{- 11 \pm 19}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 11 \pm 19}{6}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{4}{3}, - 5$