Y=4x^2-20x+4

Solution for Y=4x^2-20x+4 equation:

=4Y^2-20Y+4

We move all terms to the left:

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

We get rid of parentheses

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

a = -4; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·(-4)·(-4)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{21}}{2*-4}=\frac{-20-4\sqrt{21}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{21}}{2*-4}=\frac{-20+4\sqrt{21}}{-8}
$