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(2x^2+2x-12)/(x-2)=x
We move all terms to the left:
(2x^2+2x-12)/(x-2)-(x)=0
Domain of the equation: (x-2)!=0We add all the numbers together, and all the variables
We move all terms containing x to the left, all other terms to the right
x!=2
x∈R
-1x+(2x^2+2x-12)/(x-2)=0
We multiply all the terms by the denominator
-1x*(x-2)+(2x^2+2x-12)=0
We multiply parentheses
-x^2+2x+(2x^2+2x-12)=0
We get rid of parentheses
-x^2+2x^2+2x+2x-12=0
We add all the numbers together, and all the variables
x^2+4x-12=0
a = 1; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·1·(-12)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*1}=\frac{-12}{2} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*1}=\frac{4}{2} =2 $
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