4x^2+80x+1=0

Solution for 4x^2+80x+1=0 equation:

4x^2+80x+1=0

a = 4; b = 80; c = +1;
Δ = b2-4ac
Δ = 802-4·4·1
Δ = 6384
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:

$\sqrt{\Delta}=\sqrt{6384}=\sqrt{16*399}=\sqrt{16}*\sqrt{399}=4\sqrt{399}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-4\sqrt{399}}{2*4}=\frac{-80-4\sqrt{399}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+4\sqrt{399}}{2*4}=\frac{-80+4\sqrt{399}}{8}
$