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15t-1.86t^2=0
a = -1.86; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-1.86)·0
Δ = 225
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{225}=15$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-1.86}=\frac{-30}{-3.72} =8+0.24/3.72 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-1.86}=\frac{0}{-3.72} =0 $
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