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3t^2-7t+4=0
a = 3; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·3·4
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*3}=\frac{6}{6} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*3}=\frac{8}{6} =1+1/3 $
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