(x-4)(x+1)=-6

Solution for (x-4)(x+1)=-6 equation:

(x-4)(x+1)=-6

We move all terms to the left:

(x-4)(x+1)-(-6)=0

We add all the numbers together, and all the variables

(x-4)(x+1)+6=0

We multiply parentheses ..

(+x^2+x-4x-4)+6=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x+2=0

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

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