How do you solve #4x^2-4x=15#?

2 Answers
Sonnhard
Jun 4, 2018

#x_1=5/2# or #x_2=-3/2#

Explanation:

Writing your equation in the form
#4x^2-4x-15=0#
dividing by #4#
#x^2-x-15/4=0#
using the quadratic formula
#x_{1,2}=1/2pmsqrt(1/4+15/4)#
so we get

#x_1=5/2#

#x_2=-3/2#

Jim G.
Jun 4, 2018

#x=-3/2" or "x=5/2#

Explanation:

#"rearrange in standard form ";ax^2+bx+c=0#

#"subtract 15 from both sides"#

#4x^2-4x-15=0#

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

#"the factors of the product "4xx-15=-60#

#"which sum to - 4 are + 6 and - 10"#

#"split the middle term using these factors"#

#4x^2+6x-10x-15=0larrcolor(blue)"factor by grouping"#

#color(red)(2x)(2x+3)color(red)(-5)(2x+3)=0#

#"take out the "color(blue)"common factor "(2x+3)#

#(2x+3)(color(red)(2x-5))=0#

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

#2x+3=0rArrx=-3/2#

#2x-5=0rArrx=5/2#

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
3865 views around the world
You can reuse this answer
Creative Commons License