36=((2x+2)(3x))/2

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Solution for 36=((2x+2)(3x))/2 equation: 



36=((2x+2)(3x))/2

We move all terms to the left:

36-(((2x+2)(3x))/2)=0

We multiply all the terms by the denominator


-(((2x+2)3x)+36*2)=0



We calculate terms in parentheses: -(((2x+2)3x)+36*2), so:

((2x+2)3x)+36*2

We add all the numbers together, and all the variables

((2x+2)3x)+72



We calculate terms in parentheses: +((2x+2)3x), so:

(2x+2)3x

We multiply parentheses

6x^2+6x

Back to the equation:

+(6x^2+6x)



We get rid of parentheses

6x^2+6x+72

Back to the equation:

-(6x^2+6x+72)



We get rid of parentheses

-6x^2-6x-72=0

a = -6; b = -6; c = -72;
Δ = b2-4ac
Δ = -62-4·(-6)·(-72)
Δ = -1692
Delta is less than zero, so there is no solution for the equation

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