-26t^2+29t+6=0

Solution for -26t^2+29t+6=0 equation:

-26t^2+29t+6=0

a = -26; b = 29; c = +6;
Δ = b2-4ac
Δ = 292-4·(-26)·6
Δ = 1465
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{-(29)-\sqrt{1465}}{2*-26}=\frac{-29-\sqrt{1465}}{-52}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1465}}{2*-26}=\frac{-29+\sqrt{1465}}{-52}
$