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1.05t^2-3.7t-7.4=0
a = 1.05; b = -3.7; c = -7.4;
Δ = b2-4ac
Δ = -3.72-4·1.05·(-7.4)
Δ = 44.77
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{-(-3.7)-\sqrt{44.77}}{2*1.05}=\frac{3.7-\sqrt{44.77}}{2.1} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.7)+\sqrt{44.77}}{2*1.05}=\frac{3.7+\sqrt{44.77}}{2.1} $
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