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5t^2-11t+6=0
a = 5; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·5·6
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*5}=\frac{10}{10} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*5}=\frac{12}{10} =1+1/5 $
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