5-7k=-k(k+1)-3

Solution for 5-7k=-k(k+1)-3 equation:

5-7k=-k(k+1)-3

We move all terms to the left:

5-7k-(-k(k+1)-3)=0


We calculate terms in parentheses: -(-k(k+1)-3), so:

-k(k+1)-3

We multiply parentheses

-k^2-k-3

We add all the numbers together, and all the variables

-1k^2-1k-3

Back to the equation:

-(-1k^2-1k-3)


We get rid of parentheses

1k^2+1k-7k+3+5=0

We add all the numbers together, and all the variables

k^2-6k+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
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{-(-6)-2}{2*1}=\frac{4}{2}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2}
=4
$