20y+7y^2=32

Solution for 20y+7y^2=32 equation:

20y+7y^2=32

We move all terms to the left:

20y+7y^2-(32)=0

a = 7; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·7·(-32)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-36}{2*7}=\frac{-56}{14}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*7}=\frac{16}{14}
=1+1/7
$