(4x^2)+56x+1=0

Solution for (4x^2)+56x+1=0 equation:

(4x^2)+56x+1=0

a = 4; b = 56; c = +1;
Δ = b2-4ac
Δ = 562-4·4·1
Δ = 3120
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:

$\sqrt{\Delta}=\sqrt{3120}=\sqrt{16*195}=\sqrt{16}*\sqrt{195}=4\sqrt{195}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{195}}{2*4}=\frac{-56-4\sqrt{195}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{195}}{2*4}=\frac{-56+4\sqrt{195}}{8}
$