((-3x^2)+22x+32)/((6x^2)+12x+6)=0

Solution for ((-3x^2)+22x+32)/((6x^2)+12x+6)=0 equation:

((-3x^2)+22x+32)/((6x^2)+12x+6)=0


Domain of the equation: (6x^2+12x+6)!=0

We move all terms containing x to the left, all other terms to the right

6x^2+12x!=-6

x∈R


We multiply all the terms by the denominator


((-3x^2)+22x+32)=0


We calculate terms in parentheses: +((-3x^2)+22x+32), so:

(-3x^2)+22x+32

We get rid of parentheses

-3x^2+22x+32

Back to the equation:

+(-3x^2+22x+32)


We get rid of parentheses

-3x^2+22x+32=0

a = -3; b = 22; c = +32;
Δ = b2-4ac
Δ = 222-4·(-3)·32
Δ = 868
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{217}}{2*-3}=\frac{-22-2\sqrt{217}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{217}}{2*-3}=\frac{-22+2\sqrt{217}}{-6}
$