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4t^2-24t+9=0
a = 4; b = -24; c = +9;
Δ = b2-4ac
Δ = -242-4·4·9
Δ = 432
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{3}}{2*4}=\frac{24-12\sqrt{3}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{3}}{2*4}=\frac{24+12\sqrt{3}}{8} $
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