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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+y^2-1y+4y-4)-6=0

We get rid of parentheses

y^2-1y+4y-4-6=0

We add all the numbers together, and all the variables

y^2+3y-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$