(4x+3)(2x-5)=(4x)(2x)+(4x)(-5)+(3)(2x)+(3)(-5)=8x2-20x+6x-15=8x2-14x-15

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Solution for (4x+3)(2x-5)=(4x)(2x)+(4x)(-5)+(3)(2x)+(3)(-5)=8x2-20x+6x-15=8x2-14x-15 equation: 



(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:

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)



We get rid of parentheses

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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