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5t^2-14.14t+6=0
a = 5; b = -14.14; c = +6;
Δ = b2-4ac
Δ = -14.142-4·5·6
Δ = 79.9396
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{-(-14.14)-\sqrt{79.9396}}{2*5}=\frac{14.14-\sqrt{79.9396}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14.14)+\sqrt{79.9396}}{2*5}=\frac{14.14+\sqrt{79.9396}}{10} $
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