# How do you solve 6x² + 6x - 21 = 0?

Aug 1, 2015

${x}_{1 , 2} = - \frac{1}{2} \pm \frac{\sqrt{15}}{2}$

#### Explanation:

You could try solving this quadratic equation by completing the square.

The first thing that you need to do is get your quadratic to the form

color(blue)(x^2 + b/ax = -c/a

You can do that by adding $21$ to both sides of the equation and dividing everything by $6$.

$6 {x}^{2} + 6 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} = 21$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} + \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} = \frac{21}{6}$

${x}^{2} + x = \frac{7}{2}$

Next, use the coefficient of the $x$-term to help you determine what number must be added to both sides of the equation in order to allow for the left side of the equation to be written as the square of a binomial.

More specifically, divide this coefficient by $2$, then square the result.

${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$

So, add $\frac{1}{4}$ to both sides of the equation to get

${x}^{2} + x + \frac{1}{4} = \frac{7}{2} + \frac{1}{4}$

Now the left side of the equation can be written as

${x}^{2} + 2 \cdot \left(\frac{1}{2}\right) \cdot x + {\left(\frac{1}{2}\right)}^{2} = {\left(x + \frac{1}{2}\right)}^{2}$

The quadratic equation will be equivalent to

${\left(x + \frac{1}{2}\right)}^{2} = \frac{15}{4}$

Take the square root of both sides of the equation to get

$\sqrt{{\left(x + \frac{1}{2}\right)}^{2}} = \sqrt{\frac{15}{4}}$

$x + \frac{1}{2} = \pm \frac{\sqrt{15}}{2} \implies {x}_{1 , 2} = - \frac{1}{2} \pm \frac{\sqrt{15}}{2}$

${x}_{1} = \textcolor{g r e e n}{- \frac{1}{2} - \frac{\sqrt{15}}{2}}$ and ${x}_{2} = \textcolor{g r e e n}{- \frac{1}{2} + \frac{\sqrt{15}}{2}}$