20y^2+11y=4

Solution for 20y^2+11y=4 equation:

20y^2+11y=4

We move all terms to the left:

20y^2+11y-(4)=0

a = 20; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·20·(-4)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*20}=\frac{-32}{40}
=-4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*20}=\frac{10}{40}
=1/4
$