52x^2+40x+1=0

Solution for 52x^2+40x+1=0 equation:

52x^2+40x+1=0

a = 52; b = 40; c = +1;
Δ = b2-4ac
Δ = 402-4·52·1
Δ = 1392
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:

$\sqrt{\Delta}=\sqrt{1392}=\sqrt{16*87}=\sqrt{16}*\sqrt{87}=4\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{87}}{2*52}=\frac{-40-4\sqrt{87}}{104}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{87}}{2*52}=\frac{-40+4\sqrt{87}}{104}
$