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(2x+5)(x+3)=2^2+7x-5
We move all terms to the left:
(2x+5)(x+3)-(2^2+7x-5)=0
We get rid of parentheses
(2x+5)(x+3)-7x+5-2^2=0
We multiply parentheses ..
(+2x^2+6x+5x+15)-7x+5-2^2=0
We add all the numbers together, and all the variables
(+2x^2+6x+5x+15)-7x+1=0
We get rid of parentheses
2x^2+6x+5x-7x+15+1=0
We add all the numbers together, and all the variables
2x^2+4x+16=0
a = 2; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·2·16
Δ = -112
Delta is less than zero, so there is no solution for the equation
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