# How do you solve 4x^2 = 12x + 40?

Aug 1, 2015

${x}_{1 , 2} = \frac{3 \pm 7}{2}$

#### Explanation:

For a quadratic equation that has the general form

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

the quadratic formula takes the form

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

So, in order to be able to use this formula, get your quadratic to the general form by adding (-12x - 40 to both sides of the equation

$4 {x}^{2} - 12 x - 40 = \textcolor{red}{\cancel{\textcolor{b l a c k}{12 x}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{40}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{12 x}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{40}}}$

$4 {x}^{2} - 12 x - 40 = 0$

You can simplify this by dividing all the terms by $4$ to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} - \frac{12}{4} x - \frac{40}{4} = 0$

${x}^{2} - 3 x - 10 = 0$

The two solutions of the equation will thus be

${x}_{1 , 2} = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 10\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2} = \left\{\begin{matrix}{x}_{1} = \frac{3 + 7}{2} = \textcolor{g r e e n}{5} \\ {x}_{2} = \frac{3 - 7}{2} = \textcolor{g r e e n}{- 2}\end{matrix}\right.$

Aug 1, 2015

Solve $y = 4 {x}^{2} - 12 x - 40 = 0$
$y = 4 \left({x}^{2} - 3 x - 10\right) = 0$