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15t^2+11t+2=0
a = 15; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·15·2
Δ = 1
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{1}=1$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*15}=\frac{-12}{30} =-2/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*15}=\frac{-10}{30} =-1/3 $
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