k2+5=21.k=4

Solution for k2+5=21.k=4 equation:

k2+5=21.k=4

We move all terms to the left:

k2+5-(21.k)=0

We add all the numbers together, and all the variables

k2-(+21.k)+5=0

We add all the numbers together, and all the variables

k^2-(+21.k)+5=0

We get rid of parentheses

k^2-21.k+5=0

We add all the numbers together, and all the variables

k^2-21k+5=0

a = 1; b = -21; c = +5;
Δ = b2-4ac
Δ = -212-4·1·5
Δ = 421
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{421}}{2*1}=\frac{21-\sqrt{421}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{421}}{2*1}=\frac{21+\sqrt{421}}{2}
$