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

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

X^2-4(X+1)=2-3X

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

X^2-4X+3X-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
$