Y=r^2+12r+32

Solution for Y=r^2+12r+32 equation:

=Y^2+12Y+32

We move all terms to the left:

-(Y^2+12Y+32)=0

We get rid of parentheses

-Y^2-12Y-32=0

We add all the numbers together, and all the variables

-1Y^2-12Y-32=0

a = -1; b = -12; c = -32;
Δ = b2-4ac
Δ = -122-4·(-1)·(-32)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*-1}=\frac{8}{-2}
=-4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*-1}=\frac{16}{-2}
=-8
$