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5t^2-15t+2=0
a = 5; b = -15; c = +2;
Δ = b2-4ac
Δ = -152-4·5·2
Δ = 185
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{185}}{2*5}=\frac{15-\sqrt{185}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{185}}{2*5}=\frac{15+\sqrt{185}}{10} $
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