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g2+2g=7
We move all terms to the left:
g2+2g-(7)=0
We add all the numbers together, and all the variables
g^2+2g-7=0
a = 1; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·1·(-7)
Δ = 32
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{2}}{2*1}=\frac{-2-4\sqrt{2}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{2}}{2*1}=\frac{-2+4\sqrt{2}}{2} $
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