Y=x^2+12x-28

Solution for Y=x^2+12x-28 equation:

=Y^2+12Y-28

We move all terms to the left:

-(Y^2+12Y-28)=0

We get rid of parentheses

-Y^2-12Y+28=0

We add all the numbers together, and all the variables

-1Y^2-12Y+28=0

a = -1; b = -12; c = +28;
Δ = b2-4ac
Δ = -122-4·(-1)·28
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*-1}=\frac{-4}{-2}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*-1}=\frac{28}{-2}
=-14
$