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1.5t^2-50t-500=0
a = 1.5; b = -50; c = -500;
Δ = b2-4ac
Δ = -502-4·1.5·(-500)
Δ = 5500
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5500}=\sqrt{100*55}=\sqrt{100}*\sqrt{55}=10\sqrt{55}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{55}}{2*1.5}=\frac{50-10\sqrt{55}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{55}}{2*1.5}=\frac{50+10\sqrt{55}}{3} $
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