Question 132944
{{{4x+x(x-3)=5}}}


In order to solve a quadratic, it needs to be in standard form ({{{ax^2+bx+c=0}}}, so:


Distribute the x:
{{{4x+x^2-3x=5}}}


Collect like terms:
{{{x+x^2=5}}}


Add -5 to both sides:
{{{x+x^2-5=0}}}


And rearrange the terms in decending degree order:
{{{x^2+x-5=0}}}


This quadratic does not factor to integer factors, so let's try completing the square.


Move the constant term back to the right side:
{{{x^2+x=5}}}


Divide the coefficient on the first degree term by 2: {{{1/2}}}, square it, and add the result to both sides:
{{{x^2+x+1/4=21/4}}}


Now the left side is a perfect square, so factor it:
{{{(x+1/2)^2=21/4}}}


Take the square root of both sides:
{{{x+1/2=sqrt(21/4)}}} or {{{x+1/2=-sqrt(21/4)}}}


{{{x=-1/2+-sqrt(21/4)}}} => {{{x=(-1+-sqrt(21))/2}}}