E=(3x+2)(3x+2)+(3x+2)(3x+2)

Solution for E=(3x+2)(3x+2)+(3x+2)(3x+2) equation:

=(3E+2)(3E+2)+(3E+2)(3E+2)

We move all terms to the left:

-((3E+2)(3E+2)+(3E+2)(3E+2))=0

We multiply parentheses ..

-((+9E^2+6E+6E+4)+(+9E^2+6E+6E+4))=0


We calculate terms in parentheses: -((+9E^2+6E+6E+4)+(+9E^2+6E+6E+4)), so:

(+9E^2+6E+6E+4)+(+9E^2+6E+6E+4)

We get rid of parentheses

9E^2+9E^2+6E+6E+6E+6E+4+4

We add all the numbers together, and all the variables

18E^2+24E+8

Back to the equation:

-(18E^2+24E+8)


We get rid of parentheses

-18E^2-24E-8=0

a = -18; b = -24; c = -8;
Δ = b2-4ac
Δ = -242-4·(-18)·(-8)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$E=\frac{-b}{2a}=\frac{24}{-36}=-2/3$