5w^2+24w+5=0

Solution for 5w^2+24w+5=0 equation:

5w^2+24w+5=0

a = 5; b = 24; c = +5;
Δ = b2-4ac
Δ = 242-4·5·5
Δ = 476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{119}}{2*5}=\frac{-24-2\sqrt{119}}{10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{119}}{2*5}=\frac{-24+2\sqrt{119}}{10}
$