Y=-1y(3y+5)

Solution for Y=-1y(3y+5) equation:

=-1Y(3Y+5)

We move all terms to the left:

-(-1Y(3Y+5))=0


We calculate terms in parentheses: -(-1Y(3Y+5)), so:

-1Y(3Y+5)

We multiply parentheses

-3Y^2-5Y

Back to the equation:

-(-3Y^2-5Y)


We get rid of parentheses

3Y^2+5Y=0

a = 3; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·3·0
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*3}=\frac{-10}{6}
=-1+2/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*3}=\frac{0}{6}
=0
$