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4t^2-12t-6=0
a = 4; b = -12; c = -6;
Δ = b2-4ac
Δ = -122-4·4·(-6)
Δ = 240
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{15}}{2*4}=\frac{12-4\sqrt{15}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{15}}{2*4}=\frac{12+4\sqrt{15}}{8} $
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