Y=4x(2x+3)

Solution for Y=4x(2x+3) equation:

=4Y(2Y+3)

We move all terms to the left:

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


We calculate terms in parentheses: -(4Y(2Y+3)), so:

4Y(2Y+3)

We multiply parentheses

8Y^2+12Y

Back to the equation:

-(8Y^2+12Y)


We get rid of parentheses

-8Y^2-12Y=0

a = -8; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-8)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-8}=\frac{0}{-16}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-8}=\frac{24}{-16}
=-1+1/2
$