11+32y+-3y^2=0

Solution for 11+32y+-3y^2=0 equation:

11+32y+-3y^2=0

We add all the numbers together, and all the variables

-3y^2+32y=0

a = -3; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-3)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-3}=\frac{-64}{-6}
=10+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-3}=\frac{0}{-6}
=0
$