3(10-2k)k=1/5

Solution for 3(10-2k)k=1/5 equation:

3(10-2k)k=1/5

We move all terms to the left:

3(10-2k)k-(1/5)=0

We add all the numbers together, and all the variables

3(-2k+10)k-(+1/5)=0

We multiply parentheses

-6k^2+30k-(+1/5)=0

We get rid of parentheses

-6k^2+30k-1/5=0

We multiply all the terms by the denominator


-6k^2*5+30k*5-1=0

Wy multiply elements

-30k^2+150k-1=0

a = -30; b = 150; c = -1;
Δ = b2-4ac
Δ = 1502-4·(-30)·(-1)
Δ = 22380
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:

$\sqrt{\Delta}=\sqrt{22380}=\sqrt{4*5595}=\sqrt{4}*\sqrt{5595}=2\sqrt{5595}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-2\sqrt{5595}}{2*-30}=\frac{-150-2\sqrt{5595}}{-60}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+2\sqrt{5595}}{2*-30}=\frac{-150+2\sqrt{5595}}{-60}
$