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4.2t^2-18t+7=0
a = 4.2; b = -18; c = +7;
Δ = b2-4ac
Δ = -182-4·4.2·7
Δ = 206.4
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{-(-18)-\sqrt{206.4}}{2*4.2}=\frac{18-\sqrt{206.4}}{8.4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+\sqrt{206.4}}{2*4.2}=\frac{18+\sqrt{206.4}}{8.4} $
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