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(x+2)(x+3)=18
We move all terms to the left:
(x+2)(x+3)-(18)=0
We multiply parentheses ..
(+x^2+3x+2x+6)-18=0
We get rid of parentheses
x^2+3x+2x+6-18=0
We add all the numbers together, and all the variables
x^2+5x-12=0
a = 1; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·1·(-12)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{73}}{2*1}=\frac{-5-\sqrt{73}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*1}=\frac{-5+\sqrt{73}}{2} $
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