3x2+4x-4=2x2+13x+18

Solution for 3x2+4x-4=2x2+13x+18 equation:

3x^2+4x-4=2x^2+13x+18

We move all terms to the left:

3x^2+4x-4-(2x^2+13x+18)=0

We get rid of parentheses

3x^2-2x^2+4x-13x-18-4=0

We add all the numbers together, and all the variables

x^2-9x-22=0

a = 1; b = -9; c = -22;
Δ = b2-4ac
Δ = -92-4·1·(-22)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-13}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+13}{2*1}=\frac{22}{2}
=11
$