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-4t^2+3t+80=0
a = -4; b = 3; c = +80;
Δ = b2-4ac
Δ = 32-4·(-4)·80
Δ = 1289
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{1289}}{2*-4}=\frac{-3-\sqrt{1289}}{-8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1289}}{2*-4}=\frac{-3+\sqrt{1289}}{-8} $
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