# How do you solve 3x^2-12=0 using the quadratic formula?

Aug 20, 2017

See a solution process below:

#### Explanation:

First, we can divide each side of the equation by $\textcolor{red}{3}$ to reduce the coefficients while keeping the equation balanced:

$\frac{3 {x}^{2} - 12}{\textcolor{red}{3}} = \frac{0}{\textcolor{red}{3}}$

$\frac{3 {x}^{2}}{\textcolor{red}{3}} - \frac{12}{\textcolor{red}{3}} = 0$

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

We can rewrite this expression as:

${x}^{2} + 0 x - 4 = 0$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{0}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 4}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{0} \pm \sqrt{{\textcolor{b l u e}{0}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 4}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{\pm \sqrt{0 - \left(- 16\right)}}{2}$

$x = \frac{\pm \sqrt{0 + 16}}{2}$

$x = \frac{\pm \sqrt{16}}{2}$

$x = \frac{\pm 4}{2}$

$x = \pm 2$

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Another way to solve this problem without using the quadratic formula is to add $\textcolor{red}{12}$ to each side of the equation to isolate the ${x}^{2}$ term:

$3 {x}^{2} - 12 + \textcolor{red}{12} = 0 + \textcolor{red}{12}$

$3 {x}^{2} - 0 = 12$

$3 {x}^{2} = 12$

Next, divide each side of the equation by $\textcolor{red}{3}$ to isolate the ${x}^{2}$ while keeping the equation balanced:

$\frac{3 {x}^{2}}{\textcolor{red}{3}} = \frac{12}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} {x}^{2}}{\cancel{\textcolor{red}{3}}} = 4$

${x}^{2} = 4$

Now, take the square root of each side of the equation to solve for $x$ while keeping the equation balanced. Remember the square root of a number produces a positive AND negative result:

$\sqrt{{x}^{2}} = \pm \sqrt{4}$

$x = \pm 2$

The same result as above.