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2t^2+3t-1.2=0
a = 2; b = 3; c = -1.2;
Δ = b2-4ac
Δ = 32-4·2·(-1.2)
Δ = 18.6
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)-\sqrt{18.6}}{2*2}=\frac{-3-\sqrt{18.6}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{18.6}}{2*2}=\frac{-3+\sqrt{18.6}}{4} $
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