25x^2+14x+1=0

Solution for 25x^2+14x+1=0 equation:

25x^2+14x+1=0

a = 25; b = 14; c = +1;
Δ = b2-4ac
Δ = 142-4·25·1
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{6}}{2*25}=\frac{-14-4\sqrt{6}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{6}}{2*25}=\frac{-14+4\sqrt{6}}{50}
$