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10-2x=2+x^2
We move all terms to the left:
10-2x-(2+x^2)=0
We get rid of parentheses
-x^2-2x-2+10=0
We add all the numbers together, and all the variables
-1x^2-2x+8=0
a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-1}=\frac{-4}{-2} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2} =-4 $
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