Precalculus Examples

$5 x^{2} + 9 x + 8 = 0$

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 $a = 5$ , $b = 9$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (5 \cdot 8)}}{2 \cdot 5}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 5 \cdot 8}}{2 \cdot 5}$

Step 3.1.2

Multiply $- 4 \cdot 5 \cdot 8$ .

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Step 3.1.2.1

Multiply $- 4$ by $5$ .

$x = \frac{- 9 \pm \sqrt{81 - 20 \cdot 8}}{2 \cdot 5}$

Step 3.1.2.2

Multiply $- 20$ by $8$ .

$x = \frac{- 9 \pm \sqrt{81 - 160}}{2 \cdot 5}$

Step 3.1.3

Subtract $160$ from $81$ .

$x = \frac{- 9 \pm \sqrt{- 79}}{2 \cdot 5}$

Step 3.1.4

Rewrite $- 79$ as $- 1 (79)$ .

$x = \frac{- 9 \pm \sqrt{- 1 \cdot 79}}{2 \cdot 5}$

Step 3.1.5

Rewrite $\sqrt{- 1 (79)}$ as $\sqrt{- 1} \cdot \sqrt{79}$ .

$x = \frac{- 9 \pm \sqrt{- 1} \cdot \sqrt{79}}{2 \cdot 5}$

Step 3.1.6

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

$x = \frac{- 9 \pm i \sqrt{79}}{2 \cdot 5}$

Step 3.2

Multiply $2$ by $5$ .

$x = \frac{- 9 \pm i \sqrt{79}}{10}$