3x^2+3x-4=2

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

3x^2+3x-4=2

We move all terms to the left:

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

We add all the numbers together, and all the variables

3x^2+3x-6=0

a = 3; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·3·(-6)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*3}=\frac{6}{6}
=1
$