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6t^2+30t-2=0
a = 6; b = 30; c = -2;
Δ = b2-4ac
Δ = 302-4·6·(-2)
Δ = 948
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{948}=\sqrt{4*237}=\sqrt{4}*\sqrt{237}=2\sqrt{237}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{237}}{2*6}=\frac{-30-2\sqrt{237}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{237}}{2*6}=\frac{-30+2\sqrt{237}}{12} $
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