Pre-Algebra Examples

$2 x^{2} + 9 x = - 5$

Step 1

Add $5$ to both sides of the equation.

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

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 $a = 2$ , $b = 9$ , and $c = 5$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

Step 4.1.2

Multiply $- 4 \cdot 2 \cdot 5$ .

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Step 4.1.2.1

Multiply $- 4$ by $2$ .

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

Step 4.1.2.2

Multiply $- 8$ by $5$ .

$x = \frac{- 9 \pm \sqrt{81 - 40}}{2 \cdot 2}$

Step 4.1.3

Subtract $40$ from $81$ .

$x = \frac{- 9 \pm \sqrt{41}}{2 \cdot 2}$

Step 4.2

Multiply $2$ by $2$ .

$x = \frac{- 9 \pm \sqrt{41}}{4}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{9 - \sqrt{41}}{4}, - \frac{9 + \sqrt{41}}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{9 - \sqrt{41}}{4}, - \frac{9 + \sqrt{41}}{4}$

Decimal Form:

$x = - 0.64921894 \dots, - 3.85078105 \dots$