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5t^2-2t-30=0
a = 5; b = -2; c = -30;
Δ = b2-4ac
Δ = -22-4·5·(-30)
Δ = 604
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{151}}{2*5}=\frac{2-2\sqrt{151}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{151}}{2*5}=\frac{2+2\sqrt{151}}{10} $
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