24k^2-232k+448=0

Solution for 24k^2-232k+448=0 equation:

24k^2-232k+448=0

a = 24; b = -232; c = +448;
Δ = b2-4ac
Δ = -2322-4·24·448
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-232)-104}{2*24}=\frac{128}{48}
=2+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-232)+104}{2*24}=\frac{336}{48}
=7
$