Y=4x2+20x-96

Solution for Y=4x2+20x-96 equation:

=4Y^2+20Y-96

We move all terms to the left:

-(4Y^2+20Y-96)=0

We get rid of parentheses

-4Y^2-20Y+96=0

a = -4; b = -20; c = +96;
Δ = b2-4ac
Δ = -202-4·(-4)·96
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-44}{2*-4}=\frac{-24}{-8}
=+3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+44}{2*-4}=\frac{64}{-8}
=-8
$