(k^2)+7-3k=2k+1

Solution for (k^2)+7-3k=2k+1 equation:

(k^2)+7-3k=2k+1

We move all terms to the left:

(k^2)+7-3k-(2k+1)=0

We get rid of parentheses

k^2-3k-2k-1+7=0

We add all the numbers together, and all the variables

k^2-5k+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$