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12t^2-312t+24=0
a = 12; b = -312; c = +24;
Δ = b2-4ac
Δ = -3122-4·12·24
Δ = 96192
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96192}=\sqrt{576*167}=\sqrt{576}*\sqrt{167}=24\sqrt{167}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-312)-24\sqrt{167}}{2*12}=\frac{312-24\sqrt{167}}{24} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-312)+24\sqrt{167}}{2*12}=\frac{312+24\sqrt{167}}{24} $
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