3y2=y(y-5)

Solution for 3y2=y(y-5) equation:

3y^2=y(y-5)

We move all terms to the left:

3y^2-(y(y-5))=0


We calculate terms in parentheses: -(y(y-5)), so:

y(y-5)

We multiply parentheses

y^2-5y

Back to the equation:

-(y^2-5y)


We get rid of parentheses

3y^2-y^2+5y=0

We add all the numbers together, and all the variables

2y^2+5y=0

a = 2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·2·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*2}=\frac{-10}{4}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*2}=\frac{0}{4}
=0
$