7y(4y-5)=0

Solution for 7y(4y-5)=0 equation:

7y(4y-5)=0

We multiply parentheses

28y^2-35y=0

a = 28; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·28·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*28}=\frac{0}{56}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*28}=\frac{70}{56}
=1+1/4
$