x^2-3x=4x+18

Solution for x^2-3x=4x+18 equation:

x^2-3x=4x+18

We move all terms to the left:

x^2-3x-(4x+18)=0

We get rid of parentheses

x^2-3x-4x-18=0

We add all the numbers together, and all the variables

x^2-7x-18=0

a = 1; b = -7; c = -18;
Δ = b2-4ac
Δ = -72-4·1·(-18)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*1}=\frac{18}{2}
=9
$