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64-3t-5t^2=0
a = -5; b = -3; c = +64;
Δ = b2-4ac
Δ = -32-4·(-5)·64
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*-5}=\frac{3-\sqrt{1289}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1289}}{2*-5}=\frac{3+\sqrt{1289}}{-10} $
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