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(2x+3)(2x+5)=x
We move all terms to the left:
(2x+3)(2x+5)-(x)=0
We add all the numbers together, and all the variables
-1x+(2x+3)(2x+5)=0
We multiply parentheses ..
(+4x^2+10x+6x+15)-1x=0
We get rid of parentheses
4x^2+10x+6x-1x+15=0
We add all the numbers together, and all the variables
4x^2+15x+15=0
a = 4; b = 15; c = +15;
Δ = b2-4ac
Δ = 152-4·4·15
Δ = -15
Delta is less than zero, so there is no solution for the equation
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