20y+y^2=350

Solution for 20y+y^2=350 equation:

20y+y^2=350

We move all terms to the left:

20y+y^2-(350)=0

a = 1; b = 20; c = -350;
Δ = b2-4ac
Δ = 202-4·1·(-350)
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-30\sqrt{2}}{2*1}=\frac{-20-30\sqrt{2}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+30\sqrt{2}}{2*1}=\frac{-20+30\sqrt{2}}{2}
$