64-4k^2-28k=0

Solution for 64-4k^2-28k=0 equation:

64-4k^2-28k=0

a = -4; b = -28; c = +64;
Δ = b2-4ac
Δ = -282-4·(-4)·64
Δ = 1808
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1808}=\sqrt{16*113}=\sqrt{16}*\sqrt{113}=4\sqrt{113}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{113}}{2*-4}=\frac{28-4\sqrt{113}}{-8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{113}}{2*-4}=\frac{28+4\sqrt{113}}{-8}
$