# 0.7t^2+2t-14=0

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## Solution for 0.7t^2+2t-14=0 equation:

0.7t^2+2t-14=0
a = 0.7; b = 2; c = -14;Δ = b2-4acΔ = 22-4·0.7·(-14)Δ = 43.2The delta value is higher than zero, so the equation has two solutionsWe 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{-(2)-\sqrt{43.2}}{2*0.7}=\frac{-2-\sqrt{43.2}}{1.4}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{43.2}}{2*0.7}=\frac{-2+\sqrt{43.2}}{1.4}$

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