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3t^2-12t-5=0
a = 3; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·3·(-5)
Δ = 204
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{51}}{2*3}=\frac{12-2\sqrt{51}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{51}}{2*3}=\frac{12+2\sqrt{51}}{6} $
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