25t^2+t-526=0

Solution for 25t^2+t-526=0 equation:

25t^2+t-526=0

a = 25; b = 1; c = -526;
Δ = b2-4ac
Δ = 12-4·25·(-526)
Δ = 52601
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{52601}}{2*25}=\frac{-1-\sqrt{52601}}{50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{52601}}{2*25}=\frac{-1+\sqrt{52601}}{50}
$