4k^2+24k-3=0

Solution for 4k^2+24k-3=0 equation:

4k^2+24k-3=0

a = 4; b = 24; c = -3;
Δ = b2-4ac
Δ = 242-4·4·(-3)
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{39}}{2*4}=\frac{-24-4\sqrt{39}}{8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{39}}{2*4}=\frac{-24+4\sqrt{39}}{8}
$