3y2+4y=20

Solution for 3y2+4y=20 equation:

3y^2+4y=20

We move all terms to the left:

3y^2+4y-(20)=0

a = 3; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·3·(-20)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*3}=\frac{-20}{6}
=-3+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*3}=\frac{12}{6}
=2
$