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g(2)=2(2)^2+1
We move all terms to the left:
g(2)-(2(2)^2+1)=0
We add all the numbers together, and all the variables
g^2-485=0
a = 1; b = 0; c = -485;
Δ = b2-4ac
Δ = 02-4·1·(-485)
Δ = 1940
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1940}=\sqrt{4*485}=\sqrt{4}*\sqrt{485}=2\sqrt{485}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{485}}{2*1}=\frac{0-2\sqrt{485}}{2} =-\frac{2\sqrt{485}}{2} =-\sqrt{485} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{485}}{2*1}=\frac{0+2\sqrt{485}}{2} =\frac{2\sqrt{485}}{2} =\sqrt{485} $
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