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3x+1=x^2
We move all terms to the left:
3x+1-(x^2)=0
determiningTheFunctionDomain -x^2+3x+1=0
We add all the numbers together, and all the variables
-1x^2+3x+1=0
a = -1; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·(-1)·1
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{-(3)-\sqrt{13}}{2*-1}=\frac{-3-\sqrt{13}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*-1}=\frac{-3+\sqrt{13}}{-2} $
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