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15t^2-47t-2=0
a = 15; b = -47; c = -2;
Δ = b2-4ac
Δ = -472-4·15·(-2)
Δ = 2329
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{-(-47)-\sqrt{2329}}{2*15}=\frac{47-\sqrt{2329}}{30} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2329}}{2*15}=\frac{47+\sqrt{2329}}{30} $
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