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(x+3)(x+12)=12(x+18)
We move all terms to the left:
(x+3)(x+12)-(12(x+18))=0
We multiply parentheses ..
(+x^2+12x+3x+36)-(12(x+18))=0
We calculate terms in parentheses: -(12(x+18)), so:We get rid of parentheses
12(x+18)
We multiply parentheses
12x+216
Back to the equation:
-(12x+216)
x^2+12x+3x-12x+36-216=0
We add all the numbers together, and all the variables
x^2+3x-180=0
a = 1; b = 3; c = -180;
Δ = b2-4ac
Δ = 32-4·1·(-180)
Δ = 729
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*1}=\frac{-30}{2} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*1}=\frac{24}{2} =12 $
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