24x^2+196x+32=0

Solution for 24x^2+196x+32=0 equation:

24x^2+196x+32=0

a = 24; b = 196; c = +32;
Δ = b2-4ac
Δ = 1962-4·24·32
Δ = 35344
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{35344}=188$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(196)-188}{2*24}=\frac{-384}{48}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(196)+188}{2*24}=\frac{-8}{48}
=-1/6
$