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12t^2-11t+2=0
a = 12; b = -11; c = +2;
Δ = b2-4ac
Δ = -112-4·12·2
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*12}=\frac{6}{24} =1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*12}=\frac{16}{24} =2/3 $
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