13y^2+20y+18=180

Solution for 13y^2+20y+18=180 equation:

13y^2+20y+18=180

We move all terms to the left:

13y^2+20y+18-(180)=0

We add all the numbers together, and all the variables

13y^2+20y-162=0

a = 13; b = 20; c = -162;
Δ = b2-4ac
Δ = 202-4·13·(-162)
Δ = 8824
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:

$\sqrt{\Delta}=\sqrt{8824}=\sqrt{4*2206}=\sqrt{4}*\sqrt{2206}=2\sqrt{2206}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{2206}}{2*13}=\frac{-20-2\sqrt{2206}}{26}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{2206}}{2*13}=\frac{-20+2\sqrt{2206}}{26}
$