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1.5t^2-11t+9=0
a = 1.5; b = -11; c = +9;
Δ = b2-4ac
Δ = -112-4·1.5·9
Δ = 67
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{-(-11)-\sqrt{67}}{2*1.5}=\frac{11-\sqrt{67}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{67}}{2*1.5}=\frac{11+\sqrt{67}}{3} $
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