5t^2-24t-128=0

Solution for 5t^2-24t-128=0 equation:

5t^2-24t-128=0

a = 5; b = -24; c = -128;
Δ = b2-4ac
Δ = -242-4·5·(-128)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-56}{2*5}=\frac{-32}{10}
=-3+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+56}{2*5}=\frac{80}{10}
=8
$