5x^2+24x+1=0

Solution for 5x^2+24x+1=0 equation:

5x^2+24x+1=0

a = 5; b = 24; c = +1;
Δ = b2-4ac
Δ = 242-4·5·1
Δ = 556
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:

$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{139}}{2*5}=\frac{-24-2\sqrt{139}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{139}}{2*5}=\frac{-24+2\sqrt{139}}{10}
$