Finite Math Examples

$2 p x^{2} - 17 = - 5 q x$

Step 1

Add $5 q x$ to both sides of the equation.

$2 p x^{2} - 17 + 5 q x = 0$

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 $a = 2 p$ , $b = 5 q$ , and $c = - 17$ into the quadratic formula and solve for $x$ .

$\frac{- 5 q \pm \sqrt{{(5 q)}^{2} - 4 \cdot (2 p \cdot - 17)}}{2 (2 p)}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the product rule to $5 q$ .

$x = \frac{- 5 q \pm \sqrt{5^{2} q^{2} - 4 \cdot (2 p) \cdot - 17}}{2 \cdot (2 p)}$

Step 4.1.2

Raise $5$ to the power of $2$ .

$x = \frac{- 5 q \pm \sqrt{25 q^{2} - 4 \cdot (2 p) \cdot - 17}}{2 \cdot (2 p)}$

Step 4.1.3

Multiply $- 4$ by $2$ .

$x = \frac{- 5 q \pm \sqrt{25 q^{2} - 8 p \cdot - 17}}{2 \cdot (2 p)}$

Step 4.1.4

Multiply $- 17$ by $- 8$ .

$x = \frac{- 5 q \pm \sqrt{25 q^{2} + 136 p}}{2 \cdot (2 p)}$

Step 4.2

Multiply $2$ by $2$ .

$x = \frac{- 5 q \pm \sqrt{25 q^{2} + 136 p}}{4 p}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{5 q - \sqrt{25 q^{2} + 136 p}}{4 p}$

$x = - \frac{5 q + \sqrt{25 q^{2} + 136 p}}{4 p}$