Y^3+7y^2-18y=0

Solution for Y^3+7y^2-18y=0 equation:

^3+7Y^2-18Y=0

We add all the numbers together, and all the variables

7Y^2-18Y=0

a = 7; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·7·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*7}=\frac{0}{14}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*7}=\frac{36}{14}
=2+4/7
$