Y=3x2+4x

Solution for Y=3x2+4x equation:

=3Y^2+4Y

We move all terms to the left:

-(3Y^2+4Y)=0

We get rid of parentheses

-3Y^2-4Y=0

a = -3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-3)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-3}=\frac{0}{-6}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-3}=\frac{8}{-6}
=-1+1/3
$