5t^2+8t-2=0

Solution for 5t^2+8t-2=0 equation:

5t^2+8t-2=0

a = 5; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·5·(-2)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{26}}{2*5}=\frac{-8-2\sqrt{26}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{26}}{2*5}=\frac{-8+2\sqrt{26}}{10}
$