Solve Using the Quadratic Formula 2k^2-40=-11k

$2 k^{2} - 40 = - 11 k$

Step 1

Add $11 k$ to both sides of the equation.

$2 k^{2} - 40 + 11 k = 0$

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 $a = 2$ , $b = 11$ , and $c = - 40$ into the quadratic formula and solve for $k$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (2 \cdot - 40)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$k = \frac{- 11 \pm \sqrt{121 - 4 \cdot 2 \cdot - 40}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 40$ .

Multiply $- 4$ by $2$ .

$k = \frac{- 11 \pm \sqrt{121 - 8 \cdot - 40}}{2 \cdot 2}$

Multiply $- 8$ by $- 40$ .

$k = \frac{- 11 \pm \sqrt{121 + 320}}{2 \cdot 2}$

Add $121$ and $320$ .

$k = \frac{- 11 \pm \sqrt{441}}{2 \cdot 2}$

Rewrite $441$ as $21^{2}$ .

$k = \frac{- 11 \pm \sqrt{21^{2}}}{2 \cdot 2}$

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

$k = \frac{- 11 \pm 21}{2 \cdot 2}$

Multiply $2$ by $2$ .

$k = \frac{- 11 \pm 21}{4}$

Step 5

The final answer is the combination of both solutions.

$k = \frac{5}{2}, - 8$