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t^2-4t-16.3=0
a = 1; b = -4; c = -16.3;
Δ = b2-4ac
Δ = -42-4·1·(-16.3)
Δ = 81.2
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{-(-4)-\sqrt{81.2}}{2*1}=\frac{4-\sqrt{81.2}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{81.2}}{2*1}=\frac{4+\sqrt{81.2}}{2} $
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