-4t^2-t+8=0

Solution for -4t^2-t+8=0 equation:

-4t^2-t+8=0

We add all the numbers together, and all the variables

-4t^2-1t+8=0

a = -4; b = -1; c = +8;
Δ = b2-4ac
Δ = -12-4·(-4)·8
Δ = 129
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{-(-1)-\sqrt{129}}{2*-4}=\frac{1-\sqrt{129}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*-4}=\frac{1+\sqrt{129}}{-8}
$