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15t^2+4t-3=0
a = 15; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·15·(-3)
Δ = 196
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{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*15}=\frac{-18}{30} =-3/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*15}=\frac{10}{30} =1/3 $
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