k^2+6=7

Solution for k^2+6=7 equation:

k^2+6=7

We move all terms to the left:

k^2+6-(7)=0

We add all the numbers together, and all the variables

k^2-1=0

a = 1; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·1·(-1)
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{-(0)-2}{2*1}=\frac{-2}{2}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*1}=\frac{2}{2}
=1
$