# How do I use the quadratic formula to solve 3x^2 - 6 = 4x?

Mar 21, 2018

$x = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$

#### Explanation:

Given:

$3 {x}^{2} - 6 = 4 x$

Subtract $4 x$ from both sides to get:

$3 {x}^{2} - 4 x - 6 = 0$

This is now in standard form:

$a {x}^{2} + b x + c = 0$

with $a = 3$, $b = - 4$ and $c = - 6$

So we can apply the quadratic formula to find:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- \left(\textcolor{b l u e}{- 4}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 4}\right)}^{2} - 4 \left(\textcolor{b l u e}{3}\right) \left(\textcolor{b l u e}{- 6}\right)}}{2 \left(\textcolor{b l u e}{3}\right)}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{16 + 72}}{6}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{88}}{6}$

$\textcolor{w h i t e}{x} = \frac{4 \pm \sqrt{{2}^{2} \cdot 22}}{6}$

$\textcolor{w h i t e}{x} = \frac{4 \pm 2 \sqrt{22}}{6}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{22}}{3}$

$\textcolor{w h i t e}{x} = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$

Bonus

Notice in the derivation above that we had a factor ${2}^{2}$ in the radicand that we brought outside the square root as $2$ and then cancelled with the $2$ in the divisor.

This always happens when $b$ is even, adding an extra simplification step or two.

In such cases we can instead think of the original quadratic as taking the form:

$a {x}^{2} + 2 \mathrm{dx} + c = 0$

Then the solutions can be expressed as:

$x = \frac{- d \pm \sqrt{{d}^{2} - a c}}{a}$

In our example $a = 3$, $d = - 2$ and $c = - 6$, so we get:

$x = \frac{- \left(\textcolor{b l u e}{- 2}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 2}\right)}^{2} - \left(\textcolor{b l u e}{3}\right) \left(\textcolor{b l u e}{- 6}\right)}}{\textcolor{b l u e}{3}}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{4 + 18}}{3}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{22}}{3}$

$\textcolor{w h i t e}{x} = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$

I don't know how widely this variant of the quadratic formula is used.

If $a = 1$ then we can simplify further to say that the solutions of:

${x}^{2} + 2 \mathrm{dx} + c = 0$

are:

$x = - d \pm \sqrt{{d}^{2} - 4 c}$