(3/4)(n+3)=9

Solution for (3/4)(n+3)=9 equation:

(3/4)(n+3)=9

We move all terms to the left:

(3/4)(n+3)-(9)=0


Domain of the equation: 4)(n+3)!=0

n∈R


We add all the numbers together, and all the variables

(+3/4)(n+3)-9=0

We multiply parentheses ..

(+3n^2+3/4*3)-9=0

We multiply all the terms by the denominator


(+3n^2+3-9*4*3)=0

We get rid of parentheses

3n^2+3-9*4*3=0

We add all the numbers together, and all the variables

3n^2-105=0

a = 3; b = 0; c = -105;
Δ = b2-4ac
Δ = 02-4·3·(-105)
Δ = 1260
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:

$\sqrt{\Delta}=\sqrt{1260}=\sqrt{36*35}=\sqrt{36}*\sqrt{35}=6\sqrt{35}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{35}}{2*3}=\frac{0-6\sqrt{35}}{6}
=-\frac{6\sqrt{35}}{6}
=-\sqrt{35}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{35}}{2*3}=\frac{0+6\sqrt{35}}{6}
=\frac{6\sqrt{35}}{6}
=\sqrt{35}
$