40y+5y^2=80

Solution for 40y+5y^2=80 equation:

40y+5y^2=80

We move all terms to the left:

40y+5y^2-(80)=0

a = 5; b = 40; c = -80;
Δ = b2-4ac
Δ = 402-4·5·(-80)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{2}}{2*5}=\frac{-40-40\sqrt{2}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{2}}{2*5}=\frac{-40+40\sqrt{2}}{10}
$