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t^2+18t+16.2=0
a = 1; b = 18; c = +16.2;
Δ = b2-4ac
Δ = 182-4·1·16.2
Δ = 259.2
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{-(18)-\sqrt{259.2}}{2*1}=\frac{-18-\sqrt{259.2}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+\sqrt{259.2}}{2*1}=\frac{-18+\sqrt{259.2}}{2} $
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