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x+2=x^2+3x-1
We move all terms to the left:
x+2-(x^2+3x-1)=0
We get rid of parentheses
-x^2+x-3x+1+2=0
We add all the numbers together, and all the variables
-1x^2-2x+3=0
a = -1; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-1)·3
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-1}=\frac{-2}{-2} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-1}=\frac{6}{-2} =-3 $
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