(8k+k^2-6)=-(-10k+7-2k^2)

Solution for (8k+k^2-6)=-(-10k+7-2k^2) equation:

(8k+k^2-6)=-(-10k+7-2k^2)

We move all terms to the left:

(8k+k^2-6)-(-(-10k+7-2k^2))=0

We get rid of parentheses

k^2-(-(-10k+7-2k^2))+8k-6=0


We calculate terms in parentheses: -(-(-10k+7-2k^2)), so:

-(-10k+7-2k^2)

We get rid of parentheses

2k^2+10k-7

Back to the equation:

-(2k^2+10k-7)


We add all the numbers together, and all the variables

k^2+8k-(2k^2+10k-7)-6=0

We get rid of parentheses

k^2-2k^2+8k-10k+7-6=0

We add all the numbers together, and all the variables

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

a = -1; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-1)·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*-1}=\frac{2-2\sqrt{2}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*-1}=\frac{2+2\sqrt{2}}{-2}
$