(k+4)(k+7)=k+28

Solution for (k+4)(k+7)=k+28 equation:

(k+4)(k+7)=k+28

We move all terms to the left:

(k+4)(k+7)-(k+28)=0

We get rid of parentheses

(k+4)(k+7)-k-28=0

We multiply parentheses ..

(+k^2+7k+4k+28)-k-28=0

We add all the numbers together, and all the variables

(+k^2+7k+4k+28)-1k-28=0

We get rid of parentheses

k^2+7k+4k-1k+28-28=0

We add all the numbers together, and all the variables

k^2+10k=0

a = 1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1}=\frac{-20}{2}
=-10
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1}=\frac{0}{2}
=0
$