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4t^2-45t-2=0
a = 4; b = -45; c = -2;
Δ = b2-4ac
Δ = -452-4·4·(-2)
Δ = 2057
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2057}=\sqrt{121*17}=\sqrt{121}*\sqrt{17}=11\sqrt{17}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-11\sqrt{17}}{2*4}=\frac{45-11\sqrt{17}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+11\sqrt{17}}{2*4}=\frac{45+11\sqrt{17}}{8} $
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