4x^2+3x-48=3x^2+x

Solution for 4x^2+3x-48=3x^2+x equation:

4x^2+3x-48=3x^2+x

We move all terms to the left:

4x^2+3x-48-(3x^2+x)=0

We get rid of parentheses

4x^2-3x^2+3x-x-48=0

We add all the numbers together, and all the variables

x^2+2x-48=0

a = 1; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*1}=\frac{12}{2}
=6
$