5x^2+16x+1=0

Solution for 5x^2+16x+1=0 equation:

5x^2+16x+1=0

a = 5; b = 16; c = +1;
Δ = b2-4ac
Δ = 162-4·5·1
Δ = 236
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:

$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{59}}{2*5}=\frac{-16-2\sqrt{59}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{59}}{2*5}=\frac{-16+2\sqrt{59}}{10}
$