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(4x+3)(2x-5)=(4x)(2x)+(4x)(-5)+(3)(2x)+(3)(-5)=8x^2-20x+6x-15=8x^2-14x-15
We move all terms to the left:
(4x+3)(2x-5)-((4x)(2x)+(4x)(-5)+(3)(2x)+(3)(-5))=0
We multiply parentheses ..
(+8x^2-20x+6x-15)-(4x2x+4x(-5)+32x+3(-5))=0
We calculate terms in parentheses: -(4x2x+4x(-5)+32x+3(-5)), so:We get rid of parentheses
4x2x+4x(-5)+32x+3(-5)
We add all the numbers together, and all the variables
32x+4x2x+4x(-5)-15
We multiply parentheses
32x+4x2x-20x-15
We add all the numbers together, and all the variables
12x+4x2x-15
Back to the equation:
-(12x+4x2x-15)
8x^2-20x+6x-12x-4x2x-15+15=0
We add all the numbers together, and all the variables
8x^2-26x-4x2x=0
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