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6+x=x2
We move all terms to the left:
6+x-(x2)=0
We add all the numbers together, and all the variables
-1x^2+x+6=0
a = -1; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-1)·6
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-1}=\frac{-6}{-2} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-1}=\frac{4}{-2} =-2 $
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