3n+4=3(5n-4)3n-12

Solution for 3n+4=3(5n-4)3n-12 equation:

3n+4=3(5n-4)3n-12

We move all terms to the left:

3n+4-(3(5n-4)3n-12)=0


We calculate terms in parentheses: -(3(5n-4)3n-12), so:

3(5n-4)3n-12

We multiply parentheses

45n^2-36n-12

Back to the equation:

-(45n^2-36n-12)


We get rid of parentheses

-45n^2+3n+36n+12+4=0

We add all the numbers together, and all the variables

-45n^2+39n+16=0

a = -45; b = 39; c = +16;
Δ = b2-4ac
Δ = 392-4·(-45)·16
Δ = 4401
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:

$\sqrt{\Delta}=\sqrt{4401}=\sqrt{9*489}=\sqrt{9}*\sqrt{489}=3\sqrt{489}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3\sqrt{489}}{2*-45}=\frac{-39-3\sqrt{489}}{-90}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3\sqrt{489}}{2*-45}=\frac{-39+3\sqrt{489}}{-90}
$