5t^2+2t-192=0

Solution for 5t^2+2t-192=0 equation:

5t^2+2t-192=0

a = 5; b = 2; c = -192;
Δ = b2-4ac
Δ = 22-4·5·(-192)
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-62}{2*5}=\frac{-64}{10}
=-6+2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+62}{2*5}=\frac{60}{10}
=6
$