3k^2+10k=48

Solution for 3k^2+10k=48 equation:

3k^2+10k=48

We move all terms to the left:

3k^2+10k-(48)=0

a = 3; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·3·(-48)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-26}{2*3}=\frac{-36}{6}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+26}{2*3}=\frac{16}{6}
=2+2/3
$