3(2k-5)+24=k(3k+7)

Solution for 3(2k-5)+24=k(3k+7) equation:

3(2k-5)+24=k(3k+7)

We move all terms to the left:

3(2k-5)+24-(k(3k+7))=0

We multiply parentheses

6k-(k(3k+7))-15+24=0


We calculate terms in parentheses: -(k(3k+7)), so:

k(3k+7)

We multiply parentheses

3k^2+7k

Back to the equation:

-(3k^2+7k)


We add all the numbers together, and all the variables

6k-(3k^2+7k)+9=0

We get rid of parentheses

-3k^2+6k-7k+9=0

We add all the numbers together, and all the variables

-3k^2-1k+9=0

a = -3; b = -1; c = +9;
Δ = b2-4ac
Δ = -12-4·(-3)·9
Δ = 109
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{-(-1)-\sqrt{109}}{2*-3}=\frac{1-\sqrt{109}}{-6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{109}}{2*-3}=\frac{1+\sqrt{109}}{-6}
$