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g2+16g+10=0
We add all the numbers together, and all the variables
g^2+16g+10=0
a = 1; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·1·10
Δ = 216
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{6}}{2*1}=\frac{-16-6\sqrt{6}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{6}}{2*1}=\frac{-16+6\sqrt{6}}{2} $
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