5k+1=k^2-4

Solution for 5k+1=k^2-4 equation:

5k+1=k^2-4

We move all terms to the left:

5k+1-(k^2-4)=0

We get rid of parentheses

-k^2+5k+4+1=0

We add all the numbers together, and all the variables

-1k^2+5k+5=0

a = -1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-1)·5
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{5}}{2*-1}=\frac{-5-3\sqrt{5}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{5}}{2*-1}=\frac{-5+3\sqrt{5}}{-2}
$