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2=x^2+3x+2
We move all terms to the left:
2-(x^2+3x+2)=0
We get rid of parentheses
-x^2-3x-2+2=0
We add all the numbers together, and all the variables
-1x^2-3x=0
a = -1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-1)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-1}=\frac{0}{-2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-1}=\frac{6}{-2} =-3 $
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