4(k-3)+7=2k-k(k+8)

Solution for 4(k-3)+7=2k-k(k+8) equation:

4(k-3)+7=2k-k(k+8)

We move all terms to the left:

4(k-3)+7-(2k-k(k+8))=0

We multiply parentheses

4k-(2k-k(k+8))-12+7=0


We calculate terms in parentheses: -(2k-k(k+8)), so:

2k-k(k+8)

We multiply parentheses

-k^2+2k-8k

We add all the numbers together, and all the variables

-1k^2-6k

Back to the equation:

-(-1k^2-6k)


We add all the numbers together, and all the variables

-(-1k^2-6k)+4k-5=0

We get rid of parentheses

1k^2+6k+4k-5=0

We add all the numbers together, and all the variables

k^2+10k-5=0

a = 1; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·1·(-5)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{30}}{2*1}=\frac{-10-2\sqrt{30}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{30}}{2*1}=\frac{-10+2\sqrt{30}}{2}
$