5t^2+2t-0.2=0

Solution for 5t^2+2t-0.2=0 equation:

5t^2+2t-0.2=0

a = 5; b = 2; c = -0.2;
Δ = b2-4ac
Δ = 22-4·5·(-0.2)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*5}=\frac{-2-2\sqrt{2}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*5}=\frac{-2+2\sqrt{2}}{10}
$