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g^2+4g=22
We move all terms to the left:
g^2+4g-(22)=0
a = 1; b = 4; c = -22;
Δ = b2-4ac
Δ = 42-4·1·(-22)
Δ = 104
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{26}}{2*1}=\frac{-4-2\sqrt{26}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{26}}{2*1}=\frac{-4+2\sqrt{26}}{2} $
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