5.9t^2+2t-20=0

Solution for 5.9t^2+2t-20=0 equation:

5.9t^2+2t-20=0

a = 5.9; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·5.9·(-20)
Δ = 476
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:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{119}}{2*5.9}=\frac{-2-2\sqrt{119}}{11.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{119}}{2*5.9}=\frac{-2+2\sqrt{119}}{11.8}
$