7k(k+9)=9(k-5)-14

Solution for 7k(k+9)=9(k-5)-14 equation:

7k(k+9)=9(k-5)-14

We move all terms to the left:

7k(k+9)-(9(k-5)-14)=0

We multiply parentheses

7k^2+63k-(9(k-5)-14)=0


We calculate terms in parentheses: -(9(k-5)-14), so:

9(k-5)-14

We multiply parentheses

9k-45-14

We add all the numbers together, and all the variables

9k-59

Back to the equation:

-(9k-59)


We get rid of parentheses

7k^2+63k-9k+59=0

We add all the numbers together, and all the variables

7k^2+54k+59=0

a = 7; b = 54; c = +59;
Δ = b2-4ac
Δ = 542-4·7·59
Δ = 1264
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:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-4\sqrt{79}}{2*7}=\frac{-54-4\sqrt{79}}{14}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+4\sqrt{79}}{2*7}=\frac{-54+4\sqrt{79}}{14}
$