Finite Math Examples

$5 b^{2} + 6 b - 52 = 12 b + 4$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Subtract $12 b$ from both sides of the equation.

$5 b^{2} + 6 b - 52 - 12 b = 4$

Step 1.1.2

Subtract $4$ from both sides of the equation.

$5 b^{2} + 6 b - 52 - 12 b - 4 = 0$

Step 1.2

Simplify $5 b^{2} + 6 b - 52 - 12 b - 4$ .

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Step 1.2.1

Subtract $12 b$ from $6 b$ .

$5 b^{2} - 6 b - 52 - 4 = 0$

Step 1.2.2

Subtract $4$ from $- 52$ .

$5 b^{2} - 6 b - 56 = 0$

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 $a = 5$ , $b = - 6$ , and $c = - 56$ into the quadratic formula and solve for $b$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (5 \cdot - 56)}}{2 \cdot 5}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 6$ to the power of $2$ .

$b = \frac{6 \pm \sqrt{36 - 4 \cdot 5 \cdot - 56}}{2 \cdot 5}$

Step 4.1.2

Multiply $- 4 \cdot 5 \cdot - 56$ .

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Step 4.1.2.1

Multiply $- 4$ by $5$ .

$b = \frac{6 \pm \sqrt{36 - 20 \cdot - 56}}{2 \cdot 5}$

Step 4.1.2.2

Multiply $- 20$ by $- 56$ .

$b = \frac{6 \pm \sqrt{36 + 1120}}{2 \cdot 5}$

Step 4.1.3

Add $36$ and $1120$ .

$b = \frac{6 \pm \sqrt{1156}}{2 \cdot 5}$

Step 4.1.4

Rewrite $1156$ as $34^{2}$ .

$b = \frac{6 \pm \sqrt{34^{2}}}{2 \cdot 5}$

Step 4.1.5

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

$b = \frac{6 \pm 34}{2 \cdot 5}$

Step 4.2

Multiply $2$ by $5$ .

$b = \frac{6 \pm 34}{10}$

Step 4.3

Simplify $\frac{6 \pm 34}{10}$ .

$b = \frac{3 \pm 17}{5}$

Step 5

The final answer is the combination of both solutions.

$b = 4, - \frac{14}{5}$