52k^2+71k+24=0

Solution for 52k^2+71k+24=0 equation:

52k^2+71k+24=0

a = 52; b = 71; c = +24;
Δ = b2-4ac
Δ = 712-4·52·24
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(71)-7}{2*52}=\frac{-78}{104}
=-3/4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(71)+7}{2*52}=\frac{-64}{104}
=-8/13
$