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

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

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

We multiply parentheses ..

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


We calculate terms in parentheses: +((+x^2-2x+x-2)), so:

(+x^2-2x+x-2)

We get rid of parentheses

x^2-2x+x-2

We add all the numbers together, and all the variables

x^2-1x-2

Back to the equation:

+(x^2-1x-2)


We get rid of parentheses

x^2-1x-2-4=0

We add all the numbers together, and all the variables

x^2-1x-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$