How do you solve #-3x^2+5x=-2#?

1 Answer
May 11, 2018

Answer:

#x=-1/3" or "x=2#

Explanation:

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

#"add 2 to both sides"#

#rArr-3x^2+5x+2=0#

#"multiply through by "-1#

#rArr3x^2-5x-2=0#

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

#"the factors of the product "3xx-2=-6#

#"which sum to - 5 are - 6 and + 1"#

#"split the middle term using these factors"#

#3x^2-6x+x-2=0larrcolor(blue)"factor by grouping"#

#color(red)(3x)(x-2)color(red)(+1)(x-2)=0#

#"take out the "color(blue)"common factor "(x-2)#

#rArr(x-2)(color(red)(3x+1))=0#

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

#3x+1=0rArrx=-1/3#

#x-2=0rArrx=2#