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6t^2-24t+23=0
a = 6; b = -24; c = +23;
Δ = b2-4ac
Δ = -242-4·6·23
Δ = 24
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{6}}{2*6}=\frac{24-2\sqrt{6}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{6}}{2*6}=\frac{24+2\sqrt{6}}{12} $
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