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(x+10)(2x+5)=(2x-3)(x-4)+7
We move all terms to the left:
(x+10)(2x+5)-((2x-3)(x-4)+7)=0
We multiply parentheses ..
(+2x^2+5x+20x+50)-((2x-3)(x-4)+7)=0
We calculate terms in parentheses: -((2x-3)(x-4)+7), so:We get rid of parentheses
(2x-3)(x-4)+7
We multiply parentheses ..
(+2x^2-8x-3x+12)+7
We get rid of parentheses
2x^2-8x-3x+12+7
We add all the numbers together, and all the variables
2x^2-11x+19
Back to the equation:
-(2x^2-11x+19)
2x^2-2x^2+5x+20x+11x+50-19=0
We add all the numbers together, and all the variables
36x+31=0
We move all terms containing x to the left, all other terms to the right
36x=-31
x=-31/36
x=-31/36
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