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x^2+3x-18=9x+9
We move all terms to the left:
x^2+3x-18-(9x+9)=0
We get rid of parentheses
x^2+3x-9x-9-18=0
We add all the numbers together, and all the variables
x^2-6x-27=0
a = 1; b = -6; c = -27;
Δ = b2-4ac
Δ = -62-4·1·(-27)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*1}=\frac{18}{2} =9 $
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