2x(x+3)+12=6-2(3x-5)

Solution for 2x(x+3)+12=6-2(3x-5) equation:

2x(x+3)+12=6-2(3x-5)

We move all terms to the left:

2x(x+3)+12-(6-2(3x-5))=0

We multiply parentheses

2x^2+6x-(6-2(3x-5))+12=0


We calculate terms in parentheses: -(6-2(3x-5)), so:

6-2(3x-5)

determiningTheFunctionDomain
-2(3x-5)+6

We multiply parentheses

-6x+10+6

We add all the numbers together, and all the variables

-6x+16

Back to the equation:

-(-6x+16)


We get rid of parentheses

2x^2+6x+6x-16+12=0

We add all the numbers together, and all the variables

2x^2+12x-4=0

a = 2; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·2·(-4)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{11}}{2*2}=\frac{-12-4\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{11}}{2*2}=\frac{-12+4\sqrt{11}}{4}
$