2k(7-5k)+11=6k+3(k^2-1)

Solution for 2k(7-5k)+11=6k+3(k^2-1) equation:

2k(7-5k)+11=6k+3(k^2-1)

We move all terms to the left:

2k(7-5k)+11-(6k+3(k^2-1))=0

We add all the numbers together, and all the variables

2k(-5k+7)-(6k+3(k^2-1))+11=0

We multiply parentheses

-10k^2+14k-(6k+3(k^2-1))+11=0


We calculate terms in parentheses: -(6k+3(k^2-1)), so:

6k+3(k^2-1)

We multiply parentheses

3k^2+6k-3

Back to the equation:

-(3k^2+6k-3)


We get rid of parentheses

-10k^2-3k^2+14k-6k+3+11=0

We add all the numbers together, and all the variables

-13k^2+8k+14=0

a = -13; b = 8; c = +14;
Δ = b2-4ac
Δ = 82-4·(-13)·14
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{22}}{2*-13}=\frac{-8-6\sqrt{22}}{-26}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{22}}{2*-13}=\frac{-8+6\sqrt{22}}{-26}
$