How do you solve $(3x+3)(x+2)=2$ by factoring?

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Answer 1 · 2015-09-13T02:31:05

$x=(-9+sqrt33)/6, (-9-sqrt33)/6$

$(3x+3)(x+2)=2$

Expand the left side by using the foil method.

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$3x^2+6x+3x+6=2$

$3x^2+9x+6=2$

Subtract $2$ from both sides.

$3x^2+9x+4=0$

You now have a quadratic equation $ax^2+bx+c$ , where $a=3, b=9, and c=4$ .

Use the quadratic formula to solve for $x$ .

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the values for a, b, and c into the equation.

$x=(-9+-sqrt(9^2-4*3*4))/(2*3)=$

$x=(-9+-sqrt(81*-48))/6=$

$x=(-9+-sqrt(33))/6$

Separate the equation into two separate equations to find both solutions for $x$ .

$x=(-9+sqrt33)/6$

$x=(-9-sqrt33)/6$

How do you solve (3x+3)(x+2)=2(3x+3)(x+2)=2 by factoring?