40y^2=15y

Solution for 40y^2=15y equation:

40y^2=15y

We move all terms to the left:

40y^2-(15y)=0

a = 40; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·40·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*40}=\frac{0}{80}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*40}=\frac{30}{80}
=3/8
$