(3x-11)(20-2x)=(7x-4)+300

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Solution for (3x-11)(20-2x)=(7x-4)+300 equation: 



(3x-11)(20-2x)=(7x-4)+300

We move all terms to the left:

(3x-11)(20-2x)-((7x-4)+300)=0

We add all the numbers together, and all the variables

(3x-11)(-2x+20)-((7x-4)+300)=0

We multiply parentheses ..

(-6x^2+60x+22x-220)-((7x-4)+300)=0



We calculate terms in parentheses: -((7x-4)+300), so:

(7x-4)+300

We get rid of parentheses

7x-4+300

We add all the numbers together, and all the variables

7x+296

Back to the equation:

-(7x+296)



We get rid of parentheses

-6x^2+60x+22x-7x-220-296=0

We add all the numbers together, and all the variables

-6x^2+75x-516=0

a = -6; b = 75; c = -516;
Δ = b2-4ac
Δ = 752-4·(-6)·(-516)
Δ = -6759
Delta is less than zero, so there is no solution for the equation

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