4x^2+40x-1=0

Solution for 4x^2+40x-1=0 equation:

4x^2+40x-1=0

a = 4; b = 40; c = -1;
Δ = b2-4ac
Δ = 402-4·4·(-1)
Δ = 1616
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:

$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{101}}{2*4}=\frac{-40-4\sqrt{101}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{101}}{2*4}=\frac{-40+4\sqrt{101}}{8}
$