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15t^2-30t=0
a = 15; b = -30; c = 0;
Δ = b2-4ac
Δ = -302-4·15·0
Δ = 900
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{900}=30$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-30}{2*15}=\frac{0}{30} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+30}{2*15}=\frac{60}{30} =2 $
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