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4.9t^2-20t-100=0
a = 4.9; b = -20; c = -100;
Δ = b2-4ac
Δ = -202-4·4.9·(-100)
Δ = 2360
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2360}=\sqrt{4*590}=\sqrt{4}*\sqrt{590}=2\sqrt{590}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{590}}{2*4.9}=\frac{20-2\sqrt{590}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{590}}{2*4.9}=\frac{20+2\sqrt{590}}{9.8} $
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