13t^2-4t-284=0

Solution for 13t^2-4t-284=0 equation:

13t^2-4t-284=0

a = 13; b = -4; c = -284;
Δ = b2-4ac
Δ = -42-4·13·(-284)
Δ = 14784
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:

$\sqrt{\Delta}=\sqrt{14784}=\sqrt{64*231}=\sqrt{64}*\sqrt{231}=8\sqrt{231}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{231}}{2*13}=\frac{4-8\sqrt{231}}{26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{231}}{2*13}=\frac{4+8\sqrt{231}}{26}
$