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6+x=x2+3
We move all terms to the left:
6+x-(x2+3)=0
We add all the numbers together, and all the variables
-(+x^2+3)+x+6=0
We get rid of parentheses
-x^2+x-3+6=0
We add all the numbers together, and all the variables
-1x^2+x+3=0
a = -1; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-1)·3
Δ = 13
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*-1}=\frac{-1-\sqrt{13}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-1}=\frac{-1+\sqrt{13}}{-2} $
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