Find the Roots (Zeros) 16x^2=-72x-83

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Algebra

$16 x^{2} = - 72 x - 83$

Add $72 x$ to both sides of the equation.

$16 x^{2} + 72 x = - 83$

Add $83$ to both sides of the equation.

$16 x^{2} + 72 x + 83 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 16$ , $b = 72$ , and $c = 83$ into the quadratic formula and solve for $x$ .

$\frac{- 72 \pm \sqrt{72^{2} - 4 \cdot (16 \cdot 83)}}{2 \cdot 16}$

Simplify.

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Simplify the numerator.

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Raise $72$ to the power of $2$ .

$x = \frac{- 72 \pm \sqrt{5184 - 4 \cdot 16 \cdot 83}}{2 \cdot 16}$

Multiply $- 4$ by $16$ .

$x = \frac{- 72 \pm \sqrt{5184 - 64 \cdot 83}}{2 \cdot 16}$

Multiply $- 64$ by $83$ .

$x = \frac{- 72 \pm \sqrt{5184 - 5312}}{2 \cdot 16}$

Subtract $5312$ from $5184$ .

$x = \frac{- 72 \pm \sqrt{- 128}}{2 \cdot 16}$

Rewrite $- 128$ as $- 1 (128)$ .

$x = \frac{- 72 \pm \sqrt{- 1 \cdot 128}}{2 \cdot 16}$

Rewrite $\sqrt{- 1 (128)}$ as $\sqrt{- 1} \cdot \sqrt{128}$ .

$x = \frac{- 72 \pm \sqrt{- 1} \cdot \sqrt{128}}{2 \cdot 16}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 72 \pm i \cdot \sqrt{128}}{2 \cdot 16}$

Rewrite $128$ as $8^{2} \cdot 2$ .

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Factor $64$ out of $128$ .

$x = \frac{- 72 \pm i \cdot \sqrt{64 (2)}}{2 \cdot 16}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 72 \pm i \cdot \sqrt{8^{2} \cdot 2}}{2 \cdot 16}$

Pull terms out from under the radical.

$x = \frac{- 72 \pm i \cdot (8 \sqrt{2})}{2 \cdot 16}$

Move $8$ to the left of $i$ .

$x = \frac{- 72 \pm 8 i \sqrt{2}}{2 \cdot 16}$

Multiply $2$ by $16$ .

$x = \frac{- 72 \pm 8 i \sqrt{2}}{32}$

Simplify $\frac{- 72 \pm 8 i \sqrt{2}}{32}$ .

$x = \frac{- 9 \pm i \sqrt{2}}{4}$

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