Y=4x^2-40x+28

Solution for Y=4x^2-40x+28 equation:

=4Y^2-40Y+28

We move all terms to the left:

-(4Y^2-40Y+28)=0

We get rid of parentheses

-4Y^2+40Y-28=0

a = -4; b = 40; c = -28;
Δ = b2-4ac
Δ = 402-4·(-4)·(-28)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24\sqrt{2}}{2*-4}=\frac{-40-24\sqrt{2}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24\sqrt{2}}{2*-4}=\frac{-40+24\sqrt{2}}{-8}
$