Y=3x^2+10x=0

Solution for Y=3x^2+10x=0 equation:

=3Y^2+10Y=0

We move all terms to the left:

-(3Y^2+10Y)=0

We get rid of parentheses

-3Y^2-10Y=0

a = -3; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·(-3)·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*-3}=\frac{0}{-6}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*-3}=\frac{20}{-6}
=-3+1/3
$