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1.5t^2+3t-25=0
a = 1.5; b = 3; c = -25;
Δ = b2-4ac
Δ = 32-4·1.5·(-25)
Δ = 159
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{-(3)-\sqrt{159}}{2*1.5}=\frac{-3-\sqrt{159}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{159}}{2*1.5}=\frac{-3+\sqrt{159}}{3} $
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