Question

Find the value of k which will make the product of 2k - 5 and k-4 equal to value of k + 8

Answer 1

Solve this equation to get the value of $k$. 

$(2k-5)(k-4) = k+8$
$2k^{2}-8k-5k+20 = k+8$
$2k^{2}-13k+20 = k+8$
$2k^{2}-13k-k = 8-20$
$2k^{2}-14k = -12$
$2k^{2}-14k+12 = 0$
I am going to use the quadratic equation formula. 
I am using '+/-' for plus-minus sign. 
$k = \frac{-(-14)+/- \sqrt{(-14)^{2}-4(2)(12)} }{2(2)}$
So, it seems to be that there are two solutions for this equations, which means two values for $k$. 

i) $k = \frac{-(-14)+ \sqrt{(-14)^{2}-4(2)(12)} }{2(2)}$
   $k = 6$

ii) $k = \frac{-(-14)- \sqrt{(-14)^{2}-4(2)(12)} }{2(2)}$
    $k = 1$

I have shown you the solutions to these equations directly. 
Here, we have $k = 6$ and $k = 1$. 
So, I think that there two values of $k$: 
i) $k = 6$
ii) $k = 1$

Hope this may help you. 

Answer 2

(2k-5)(k-4) = k+8
⇒ 2k² - 8k -5k +20 = k+8
⇒ 2k² - 13k + 20 = k + 8
⇒ 2k² - 13k + 20 - k - 8 = 0
⇒ 2k² - 14k + 12 = 0
⇒ k² - 7k + 6 = 0
⇒ k² -6k - k + 6 = 0
⇒ k(k - 6) - 1(k - 6) = 0
⇒ (k - 1)(k - 6) = 0
Therefore
k-1 = 0 ⇒ k = 1              or,                   k - 6 = 0 ⇒ k = 6