-.6t^2+6t+5=0

Solution for -.6t^2+6t+5=0 equation:

-.6t^2+6t+5=0

We add all the numbers together, and all the variables

-0.6t^2+6t+5=0

a = -0.6; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·(-0.6)·5
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*-0.6}=\frac{-6-4\sqrt{3}}{-1.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*-0.6}=\frac{-6+4\sqrt{3}}{-1.2}
$