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0.375t^2+6t=0
a = 0.375; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·0.375·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*0.375}=\frac{-12}{0.75} =-16 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*0.375}=\frac{0}{0.75} =0 $
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