Y=2x^2-28x+80

Solution for Y=2x^2-28x+80 equation:

=2Y^2-28Y+80

We move all terms to the left:

-(2Y^2-28Y+80)=0

We get rid of parentheses

-2Y^2+28Y-80=0

a = -2; b = 28; c = -80;
Δ = b2-4ac
Δ = 282-4·(-2)·(-80)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*-2}=\frac{-40}{-4}
=+10
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*-2}=\frac{-16}{-4}
=+4
$