((k+5)(k+2))/2=44

Solution for ((k+5)(k+2))/2=44 equation:

((k+5)(k+2))/2=44

We move all terms to the left:

((k+5)(k+2))/2-(44)=0

We multiply parentheses ..

((+k^2+2k+5k+10))/2-44=0

We multiply all the terms by the denominator


((+k^2+2k+5k+10))-44*2=0


We calculate terms in parentheses: +((+k^2+2k+5k+10)), so:

(+k^2+2k+5k+10)

We get rid of parentheses

k^2+2k+5k+10

We add all the numbers together, and all the variables

k^2+7k+10

Back to the equation:

+(k^2+7k+10)


We add all the numbers together, and all the variables

(k^2+7k+10)-88=0

We get rid of parentheses

k^2+7k+10-88=0

We add all the numbers together, and all the variables

k^2+7k-78=0

a = 1; b = 7; c = -78;
Δ = b2-4ac
Δ = 72-4·1·(-78)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-19}{2*1}=\frac{-26}{2}
=-13
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+19}{2*1}=\frac{12}{2}
=6
$