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.75t^2+45t-160=0
a = .75; b = 45; c = -160;
Δ = b2-4ac
Δ = 452-4·.75·(-160)
Δ = 2505
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{-(45)-\sqrt{2505}}{2*.75}=\frac{-45-\sqrt{2505}}{1.5} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2505}}{2*.75}=\frac{-45+\sqrt{2505}}{1.5} $
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