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6=x^2+x
We move all terms to the left:
6-(x^2+x)=0
We get rid of parentheses
-x^2-x+6=0
We add all the numbers together, and all the variables
-1x^2-1x+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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-1}=\frac{-4}{-2} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-1}=\frac{6}{-2} =-3 $
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