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-4+-1t+3t2=0

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Solution for -4+-1t+3t2=0 equation:



-4+-1t+3t^2=0
We add all the numbers together, and all the variables
3t^2-1t=0
a = 3; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·3·0
Δ = 1
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}

\sqrt{\Delta}=\sqrt{1}=1
t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*3}=\frac{0}{6} =0
t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*3}=\frac{2}{6} =1/3

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