# How do you solve for #x# in this problem? #4x^2 - 3 = -x#?

##### 3 Answers

#### Explanation:

#"rearrange into standard form "color(white)(x)ax^2+bx+c=0#

#"add "x" to both sides"#

#rArr4x^2+x-3=0larrcolor(blue)"in standard form"#

#"factor the quadratic using the a-c method"#

#"the factors of the product "4xx-3=-12#

#"which sum to + 1 are + 4 and - 3"#

#"split the middle term using these factors"#

#4x^2+4x-3x-3=0larrcolor(blue)"factor by grouping"#

#color(red)(4x)(x+1)color(red)(-3)(x+1)=0#

#"take out the "color(blue)"common factor "(x+1)#

#rArr(x+1)(color(red)(4x-3))=0#

#"equate each factor to zero and solve for x"#

#x+1=0rArrx=-1#

#4x-3=0rArrx=3/4#

Use the quadratic formula

#### Explanation:

First we rearrange;

now we have

and

After subbing everything in we get

the answer

#### Explanation:

show below