Y=2x^2+28x+96

Solution for Y=2x^2+28x+96 equation:

=2Y^2+28Y+96

We move all terms to the left:

-(2Y^2+28Y+96)=0

We get rid of parentheses

-2Y^2-28Y-96=0

a = -2; b = -28; c = -96;
Δ = b2-4ac
Δ = -282-4·(-2)·(-96)
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{-(-28)-4}{2*-2}=\frac{24}{-4}
=-6
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*-2}=\frac{32}{-4}
=-8
$