-3n(n-5)=3(1-3n)

Solution for -3n(n-5)=3(1-3n) equation:

-3n(n-5)=3(1-3n)

We move all terms to the left:

-3n(n-5)-(3(1-3n))=0

We add all the numbers together, and all the variables

-3n(n-5)-(3(-3n+1))=0

We multiply parentheses

-3n^2+15n-(3(-3n+1))=0


We calculate terms in parentheses: -(3(-3n+1)), so:

3(-3n+1)

We multiply parentheses

-9n+3

Back to the equation:

-(-9n+3)


We get rid of parentheses

-3n^2+15n+9n-3=0

We add all the numbers together, and all the variables

-3n^2+24n-3=0

a = -3; b = 24; c = -3;
Δ = b2-4ac
Δ = 242-4·(-3)·(-3)
Δ = 540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{15}}{2*-3}=\frac{-24-6\sqrt{15}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{15}}{2*-3}=\frac{-24+6\sqrt{15}}{-6}
$