Y=4x2+40x+5

Solution for Y=4x2+40x+5 equation:

=4Y^2+40Y+5

We move all terms to the left:

-(4Y^2+40Y+5)=0

We get rid of parentheses

-4Y^2-40Y-5=0

a = -4; b = -40; c = -5;
Δ = b2-4ac
Δ = -402-4·(-4)·(-5)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{95}}{2*-4}=\frac{40-4\sqrt{95}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{95}}{2*-4}=\frac{40+4\sqrt{95}}{-8}
$