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r2+3r=-2
We move all terms to the left:
r2+3r-(-2)=0
We add all the numbers together, and all the variables
r^2+3r+2=0
a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*1}=\frac{-4}{2} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*1}=\frac{-2}{2} =-1 $
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