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-2t^2-36t-100=0
a = -2; b = -36; c = -100;
Δ = b2-4ac
Δ = -362-4·(-2)·(-100)
Δ = 496
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{31}}{2*-2}=\frac{36-4\sqrt{31}}{-4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{31}}{2*-2}=\frac{36+4\sqrt{31}}{-4} $
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