Y=(c-5)(c+1)

Solution for Y=(c-5)(c+1) equation:

=(Y-5)(Y+1)

We move all terms to the left:

-((Y-5)(Y+1))=0

We multiply parentheses ..

-((+Y^2+Y-5Y-5))=0


We calculate terms in parentheses: -((+Y^2+Y-5Y-5)), so:

(+Y^2+Y-5Y-5)

We get rid of parentheses

Y^2+Y-5Y-5

We add all the numbers together, and all the variables

Y^2-4Y-5

Back to the equation:

-(Y^2-4Y-5)


We get rid of parentheses

-Y^2+4Y+5=0

We add all the numbers together, and all the variables

-1Y^2+4Y+5=0

a = -1; b = 4; c = +5;
Δ = b2-4ac
Δ = 42-4·(-1)·5
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*-1}=\frac{-10}{-2}
=+5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*-1}=\frac{2}{-2}
=-1
$