Finite Math Examples

$0 = - 0.5 x^{2} + 1.5 x + 9$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Add $0.5 x^{2}$ to both sides of the equation.

$0.5 x^{2} = 1.5 x + 9$

Step 1.2

Subtract $1.5 x$ from both sides of the equation.

$0.5 x^{2} - 1.5 x = 9$

Step 1.3

Subtract $9$ from both sides of the equation.

$0.5 x^{2} - 1.5 x - 9 = 0$

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 $a = 0.5$ , $b = - 1.5$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{1.5 \pm \sqrt{{(- 1.5)}^{2} - 4 \cdot (0.5 \cdot - 9)}}{2 \cdot 0.5}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 1.5$ to the power of $2$ .

$x = \frac{1.5 \pm \sqrt{2.25 - 4 \cdot 0.5 \cdot - 9}}{2 \cdot 0.5}$

Step 4.1.2

Multiply $- 4 \cdot 0.5 \cdot - 9$ .

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Step 4.1.2.1

Multiply $- 4$ by $0.5$ .

$x = \frac{1.5 \pm \sqrt{2.25 - 2 \cdot - 9}}{2 \cdot 0.5}$

Step 4.1.2.2

Multiply $- 2$ by $- 9$ .

$x = \frac{1.5 \pm \sqrt{2.25 + 18}}{2 \cdot 0.5}$

Step 4.1.3

Add $2.25$ and $18$ .

$x = \frac{1.5 \pm \sqrt{20.25}}{2 \cdot 0.5}$

Step 4.1.4

Rewrite $20.25$ as ${4.5}^{2}$ .

$x = \frac{1.5 \pm \sqrt{{4.5}^{2}}}{2 \cdot 0.5}$

Step 4.1.5

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

$x = \frac{1.5 \pm 4.5}{2 \cdot 0.5}$

Step 4.2

Multiply $2$ by $0.5$ .

$x = \frac{1.5 \pm 4.5}{1}$

Step 4.3

Divide $1.5 \pm 4.5$ by $1$ .

$x = 1.5 \pm 4.5$

Step 5

The final answer is the combination of both solutions.

$x = 6, - 3$