6k-2k(k-5)=-4+2k

Solution for 6k-2k(k-5)=-4+2k equation:

6k-2k(k-5)=-4+2k

We move all terms to the left:

6k-2k(k-5)-(-4+2k)=0

We add all the numbers together, and all the variables

6k-2k(k-5)-(2k-4)=0

We multiply parentheses

-2k^2+6k+10k-(2k-4)=0

We get rid of parentheses

-2k^2+6k+10k-2k+4=0

We add all the numbers together, and all the variables

-2k^2+14k+4=0

a = -2; b = 14; c = +4;
Δ = b2-4ac
Δ = 142-4·(-2)·4
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{57}}{2*-2}=\frac{-14-2\sqrt{57}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{57}}{2*-2}=\frac{-14+2\sqrt{57}}{-4}
$