Y=3x^2+4x*2

Solution for Y=3x^2+4x*2 equation:

=3Y^2+4Y*2

We move all terms to the left:

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

We get rid of parentheses

-3Y^2-4Y*2=0

Wy multiply elements

-3Y^2-8Y=0

a = -3; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-3)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-3}=\frac{0}{-6}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-3}=\frac{16}{-6}
=-2+2/3
$