k^2=-14k-48

Solution for k^2=-14k-48 equation:

k^2=-14k-48

We move all terms to the left:

k^2-(-14k-48)=0

We get rid of parentheses

k^2+14k+48=0

a = 1; b = 14; c = +48;
Δ = b2-4ac
Δ = 142-4·1·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*1}=\frac{-16}{2}
=-8
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*1}=\frac{-12}{2}
=-6
$