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67-5t-5t^2=0
a = -5; b = -5; c = +67;
Δ = b2-4ac
Δ = -52-4·(-5)·67
Δ = 1365
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{-(-5)-\sqrt{1365}}{2*-5}=\frac{5-\sqrt{1365}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{1365}}{2*-5}=\frac{5+\sqrt{1365}}{-10} $
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