25x^2-134x+161=0

Solution for 25x^2-134x+161=0 equation:

25x^2-134x+161=0

a = 25; b = -134; c = +161;
Δ = b2-4ac
Δ = -1342-4·25·161
Δ = 1856
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-134)-8\sqrt{29}}{2*25}=\frac{134-8\sqrt{29}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-134)+8\sqrt{29}}{2*25}=\frac{134+8\sqrt{29}}{50}
$