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g2+4g+3=0
We add all the numbers together, and all the variables
g^2+4g+3=0
a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2} =-3 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2} =-1 $
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