20y^2=7y+3

Solution for 20y^2=7y+3 equation:

20y^2=7y+3

We move all terms to the left:

20y^2-(7y+3)=0

We get rid of parentheses

20y^2-7y-3=0

a = 20; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·20·(-3)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*20}=\frac{-10}{40}
=-1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*20}=\frac{24}{40}
=3/5
$