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g^2+16g+15=0
a = 1; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·1·15
Δ = 196
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*1}=\frac{-30}{2} =-15 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*1}=\frac{-2}{2} =-1 $
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