# How do you solve 15=8x^2-14x?

Apr 22, 2016

$\textcolor{b l u e}{\textcolor{g r e e n}{x = \frac{7}{8} \pm \frac{13}{8}} = \frac{5}{2} \mathmr{and} \frac{3}{4}}$

#### Explanation:

I am opting to use vertex form equation

Write as:$\text{ } y = 8 {x}^{2} - 14 x - 15$

Compare to standard form:$\text{ } y = a {x}^{2} + b x + c$

Where$\text{ "a=8"; "b=-14"; } c = - 15$

Vertex form is:$\text{ } y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - a {\left(\frac{b}{2 a}\right)}^{2}$

$\implies y = 8 {\left(x - \frac{14}{2 \times 8}\right)}^{2} - 15 - \frac{{\left(- 14\right)}^{2}}{4 \times 8}$

$\implies y = 8 {\left(x - \frac{7}{8}\right)}^{2} - \frac{169}{8}$

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color(blue)("Vertex "->(x,y)=(+7/8,-169/8)

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Set $y = 0$ giving:

$\frac{169}{64} = {\left(x - \frac{7}{8}\right)}^{2}$

Taking the square root of both sides

$\pm \frac{13}{8} = x - \frac{7}{8}$

$\textcolor{b l u e}{\implies x = \frac{7}{8} \pm \frac{13}{8} = \frac{5}{2} \mathmr{and} \frac{3}{4}}$