5t^2-0.5t-1=0

Solution for 5t^2-0.5t-1=0 equation:

5t^2-0.5t-1=0

a = 5; b = -0.5; c = -1;
Δ = b2-4ac
Δ = -0.52-4·5·(-1)
Δ = 20.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.5)-\sqrt{20.25}}{2*5}=\frac{0.5-\sqrt{20.25}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{20.25}}{2*5}=\frac{0.5+\sqrt{20.25}}{10}
$