Y=x^2+10x-96

Solution for Y=x^2+10x-96 equation:

=Y^2+10Y-96

We move all terms to the left:

-(Y^2+10Y-96)=0

We get rid of parentheses

-Y^2-10Y+96=0

We add all the numbers together, and all the variables

-1Y^2-10Y+96=0

a = -1; b = -10; c = +96;
Δ = b2-4ac
Δ = -102-4·(-1)·96
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*-1}=\frac{-12}{-2}
=+6
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*-1}=\frac{32}{-2}
=-16
$