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4g^2+32g+60=0
a = 4; b = 32; c = +60;
Δ = b2-4ac
Δ = 322-4·4·60
Δ = 64
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{64}=8$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8}{2*4}=\frac{-40}{8} =-5 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8}{2*4}=\frac{-24}{8} =-3 $
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