k^2-6k-48=7

Solution for k^2-6k-48=7 equation:

k^2-6k-48=7

We move all terms to the left:

k^2-6k-48-(7)=0

We add all the numbers together, and all the variables

k^2-6k-55=0

a = 1; b = -6; c = -55;
Δ = b2-4ac
Δ = -62-4·1·(-55)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-16}{2*1}=\frac{-10}{2}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+16}{2*1}=\frac{22}{2}
=11
$