Solve Using the Quadratic Formula -4n^2+6n-16=-5n^2

$- 4 n^{2} + 6 n - 16 = - 5 n^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

Add $5 n^{2}$ to both sides of the equation.

$- 4 n^{2} + 6 n - 16 + 5 n^{2} = 0$

Add $- 4 n^{2}$ and $5 n^{2}$ .

$n^{2} + 6 n - 16 = 0$

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 $a = 1$ , $b = 6$ , and $c = - 16$ into the quadratic formula and solve for $n$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot - 16)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$n = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 16}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 16$ .

Multiply $- 4$ by $1$ .

$n = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 16}}{2 \cdot 1}$

Multiply $- 4$ by $- 16$ .

$n = \frac{- 6 \pm \sqrt{36 + 64}}{2 \cdot 1}$

Add $36$ and $64$ .

$n = \frac{- 6 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$n = \frac{- 6 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$n = \frac{- 6 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 6 \pm 10}{2}$

Simplify $\frac{- 6 \pm 10}{2}$ .

$n = - 3 \pm 5$

Step 5

The final answer is the combination of both solutions.

$n = 2, - 8$