# How do you solve x^2 – 3x = 4x – 1 using the quadratic formula?

Apr 17, 2016

First, we have to move everything to the left so it resembles this form: $a {x}^{2} + b x + c = 0$, also known as the standard form.

For this case, all we have to do is move everything to 1 side (it doesn't matter which side, although I prefer to move it all to the side with ${x}^{2}$ already in it)

${x}^{2} - 3 x = 4 x - 1$
${x}^{2} - 7 x + 1 = 0$

Now you will need to know the quadratic formula to solve this problem: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

But what are $a$, $b$, and $c$ you ask? Well that's why we rearranged the original equation. $a$ is the number next to ${x}^{2}$, $b$ is the number next to $x$, and $c$ doesn't have an $x$ next to it. Doing this, we find $a = 1$, $b = - 7$, $c = 1$. All we have to do is put into the quadratic formula and simplify:

$x = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$
$x = \frac{7 \pm \sqrt{49 - 4}}{2}$
$x = \frac{7 \pm \sqrt{45}}{2}$

So this gives us two answers:
$x = \frac{7 + \sqrt{45}}{2} \mathmr{and} \frac{7 - \sqrt{45}}{2}$