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1.86t^2-15t+15=0
a = 1.86; b = -15; c = +15;
Δ = b2-4ac
Δ = -152-4·1.86·15
Δ = 113.4
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{113.4}}{2*1.86}=\frac{15-\sqrt{113.4}}{3.72} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{113.4}}{2*1.86}=\frac{15+\sqrt{113.4}}{3.72} $
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