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4t^2-30t-1=0
a = 4; b = -30; c = -1;
Δ = b2-4ac
Δ = -302-4·4·(-1)
Δ = 916
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{916}=\sqrt{4*229}=\sqrt{4}*\sqrt{229}=2\sqrt{229}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{229}}{2*4}=\frac{30-2\sqrt{229}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{229}}{2*4}=\frac{30+2\sqrt{229}}{8} $
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