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-5t^2+15t+2=0
a = -5; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·(-5)·2
Δ = 265
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{265}}{2*-5}=\frac{-15-\sqrt{265}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{265}}{2*-5}=\frac{-15+\sqrt{265}}{-10} $
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