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g(2)=-3g+2
We move all terms to the left:
g(2)-(-3g+2)=0
We add all the numbers together, and all the variables
g^2-(-3g+2)=0
We get rid of parentheses
g^2+3g-2=0
a = 1; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·1·(-2)
Δ = 17
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{17}}{2*1}=\frac{-3-\sqrt{17}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17}}{2*1}=\frac{-3+\sqrt{17}}{2} $
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