Y=4x^2+20x+6

Solution for Y=4x^2+20x+6 equation:

=4Y^2+20Y+6

We move all terms to the left:

-(4Y^2+20Y+6)=0

We get rid of parentheses

-4Y^2-20Y-6=0

a = -4; b = -20; c = -6;
Δ = b2-4ac
Δ = -202-4·(-4)·(-6)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{19}}{2*-4}=\frac{20-4\sqrt{19}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{19}}{2*-4}=\frac{20+4\sqrt{19}}{-8}
$