(10-2k)k=1/5

Solution for (10-2k)k=1/5 equation:

(10-2k)k=1/5

We move all terms to the left:

(10-2k)k-(1/5)=0

We add all the numbers together, and all the variables

(-2k+10)k-(+1/5)=0

We multiply parentheses

-2k^2+10k-(+1/5)=0

We get rid of parentheses

-2k^2+10k-1/5=0

We multiply all the terms by the denominator


-2k^2*5+10k*5-1=0

Wy multiply elements

-10k^2+50k-1=0

a = -10; b = 50; c = -1;
Δ = b2-4ac
Δ = 502-4·(-10)·(-1)
Δ = 2460
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:

$\sqrt{\Delta}=\sqrt{2460}=\sqrt{4*615}=\sqrt{4}*\sqrt{615}=2\sqrt{615}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{615}}{2*-10}=\frac{-50-2\sqrt{615}}{-20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{615}}{2*-10}=\frac{-50+2\sqrt{615}}{-20}
$