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10g^2+23g+12=0
a = 10; b = 23; c = +12;
Δ = b2-4ac
Δ = 232-4·10·12
Δ = 49
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{49}=7$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*10}=\frac{-30}{20} =-1+1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*10}=\frac{-16}{20} =-4/5 $
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