k^2-4=21

Solution for k^2-4=21 equation:

k^2-4=21

We move all terms to the left:

k^2-4-(21)=0

We add all the numbers together, and all the variables

k^2-25=0

a = 1; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·1·(-25)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*1}=\frac{-10}{2}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*1}=\frac{10}{2}
=5
$