k+9=6(k-11)k=15

Solution for k+9=6(k-11)k=15 equation:

k+9=6(k-11)k=15

We move all terms to the left:

k+9-(6(k-11)k)=0


We calculate terms in parentheses: -(6(k-11)k), so:

6(k-11)k

We multiply parentheses

6k^2-66k

Back to the equation:

-(6k^2-66k)


We get rid of parentheses

-6k^2+k+66k+9=0

We add all the numbers together, and all the variables

-6k^2+67k+9=0

a = -6; b = 67; c = +9;
Δ = b2-4ac
Δ = 672-4·(-6)·9
Δ = 4705
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(67)-\sqrt{4705}}{2*-6}=\frac{-67-\sqrt{4705}}{-12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+\sqrt{4705}}{2*-6}=\frac{-67+\sqrt{4705}}{-12}
$