x^2-4+3(x+2)(x-1)=0

Solution for x^2-4+3(x+2)(x-1)=0 equation:

x^2-4+3(x+2)(x-1)=0

We multiply parentheses ..

x^2+3(+x^2-1x+2x-2)-4=0

We multiply parentheses

x^2+3x^2-3x+6x-6-4=0

We add all the numbers together, and all the variables

4x^2+3x-10=0

a = 4; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·4·(-10)
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{-(3)-13}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*4}=\frac{10}{8}
=1+1/4
$