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(4x-7)(x+3)=(2x-5)(5+2x)
We move all terms to the left:
(4x-7)(x+3)-((2x-5)(5+2x))=0
We add all the numbers together, and all the variables
(4x-7)(x+3)-((2x-5)(2x+5))=0
We use the square of the difference formula
4x^2+(4x-7)(x+3)+25=0
We multiply parentheses ..
4x^2+(+4x^2+12x-7x-21)+25=0
We get rid of parentheses
4x^2+4x^2+12x-7x-21+25=0
We add all the numbers together, and all the variables
8x^2+5x+4=0
a = 8; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·8·4
Δ = -103
Delta is less than zero, so there is no solution for the equation
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