5k^2+22k-48=0

Solution for 5k^2+22k-48=0 equation:

5k^2+22k-48=0

a = 5; b = 22; c = -48;
Δ = b2-4ac
Δ = 222-4·5·(-48)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-38}{2*5}=\frac{-60}{10}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*5}=\frac{16}{10}
=1+3/5
$