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(x+2)(x+3)=x-10
We move all terms to the left:
(x+2)(x+3)-(x-10)=0
We get rid of parentheses
(x+2)(x+3)-x+10=0
We multiply parentheses ..
(+x^2+3x+2x+6)-x+10=0
We add all the numbers together, and all the variables
(+x^2+3x+2x+6)-1x+10=0
We get rid of parentheses
x^2+3x+2x-1x+6+10=0
We add all the numbers together, and all the variables
x^2+4x+16=0
a = 1; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·1·16
Δ = -48
Delta is less than zero, so there is no solution for the equation
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