(3x^2/6x)+9=15

Solution for (3x^2/6x)+9=15 equation:

(3x^2/6x)+9=15

We move all terms to the left:

(3x^2/6x)+9-(15)=0


Domain of the equation: 6x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(3x^2/6x)-6=0

We get rid of parentheses

3x^2/6x-6=0

We multiply all the terms by the denominator


3x^2-6*6x=0

Wy multiply elements

3x^2-36x=0

a = 3; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·3·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*3}=\frac{72}{6}
=12
$