24x^2+128x+3=0

Solution for 24x^2+128x+3=0 equation:

24x^2+128x+3=0

a = 24; b = 128; c = +3;
Δ = b2-4ac
Δ = 1282-4·24·3
Δ = 16096
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:

$\sqrt{\Delta}=\sqrt{16096}=\sqrt{16*1006}=\sqrt{16}*\sqrt{1006}=4\sqrt{1006}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-4\sqrt{1006}}{2*24}=\frac{-128-4\sqrt{1006}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+4\sqrt{1006}}{2*24}=\frac{-128+4\sqrt{1006}}{48}
$