Question

X2+5x=-2x-12 solve for x

Answer 1

`x^2+5x=-2x-12\ \ \ |add\ 2x\ and\ 12\ to\ both\ sides\\\\x^2+7x+12=0\ \ \ |use\ a\ quadratic\ formula\\-----------------------\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textgreater \ 0\ then\ x_1=(-b-√(\Delta))/(2a)\ and\ x_2=(-b+√(\Delta))/(2a)\\-----------------------`


`x^2+7x+12=0\to a=1;\ b=7;\ c=12\\\\\Delta=7^2-4\cdot1\cdot12=49-48=1 \ \textgreater \ 0\ :)\ two\ solutions\\√(\Delta)}=\sqrt1=1\\\\x_1=(-7-1)/(2\cdot1)=(-8)/(2)=-4\ and\ x_2=(-7+1)/(2\cdot1)=(-6)/(2)=-3\\\\Your\ answer:\huge\boxed{x=-4\ or\ x=-3}`

Answer 2

x² + 5x = -2x -12
Add 2x to each side.
x² + 7x = -12
Add 12 to each side.
x² + 7x + 12 = 0
Let's solve this quadratic by factoring.
We want two numbers that add to get 7 and multiply to make 12. These are 3 and 4.
Split the middle of the quadratic.
x² + 3x + 4x + 12 = 0
Factor the first two and last two terms.
(You should get the same thing inside the parentheses.)
x(x+3) + 4(x+3) = 0
(x+4)(x+3) = 0
Any value of x which causes either factor to equal 0 is a solution.
We can derive the following equations and solutions:
x+4 = 0 ⇒ x = -4
x+3 = 0 ⇒ x = -3