Algebra Examples

$2 x^{2} - 3 x - 15 = 5$

Step 1

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

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Step 1.1

Subtract $5$ from both sides of the equation.

$2 x^{2} - 3 x - 15 - 5 = 0$

Step 1.2

Subtract $5$ from $- 15$ .

$2 x^{2} - 3 x - 20 = 0$

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 $a = 2$ , $b = - 3$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 20)}}{2 \cdot 2}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 20}}{2 \cdot 2}$

Step 4.1.2

Multiply $- 4 \cdot 2 \cdot - 20$ .

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Step 4.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 20}}{2 \cdot 2}$

Step 4.1.2.2

Multiply $- 8$ by $- 20$ .

$x = \frac{3 \pm \sqrt{9 + 160}}{2 \cdot 2}$

Step 4.1.3

Add $9$ and $160$ .

$x = \frac{3 \pm \sqrt{169}}{2 \cdot 2}$

Step 4.1.4

Rewrite $169$ as $13^{2}$ .

$x = \frac{3 \pm \sqrt{13^{2}}}{2 \cdot 2}$

Step 4.1.5

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

$x = \frac{3 \pm 13}{2 \cdot 2}$

Step 4.2

Multiply $2$ by $2$ .

$x = \frac{3 \pm 13}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - \frac{5}{2}$