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30t-4.9t^2+20=0
a = -4.9; b = 30; c = +20;
Δ = b2-4ac
Δ = 302-4·(-4.9)·20
Δ = 1292
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1292}=\sqrt{4*323}=\sqrt{4}*\sqrt{323}=2\sqrt{323}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{323}}{2*-4.9}=\frac{-30-2\sqrt{323}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{323}}{2*-4.9}=\frac{-30+2\sqrt{323}}{-9.8} $
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