(40y^2+10y)/5y=0

Solution for (40y^2+10y)/5y=0 equation:

(40y^2+10y)/5y=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We multiply all the terms by the denominator


(40y^2+10y)=0

We get rid of parentheses

40y^2+10y=0

a = 40; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·40·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*40}=\frac{-20}{80}
=-1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*40}=\frac{0}{80}
=0
$