Question 870120
{{{5x^2 = 8x+21}}} is a quadratic equation.
It would be customarily written in the more traditional form
{{{5x^2 - 8x-21=0}}} .
There are several ways to solve it.
All quadratic equations can be solved two ways:
by "completing the square", and
by using the quadratic formula.
Some quadratic equations can also be solved by factoring.
Each method has its advantages and disadvantages,
some equations make one method look easier,
and people have their preferences for one or another method.
 
FACTORING sometimes is very easy.
It is not too easy for this equation, but it works, and it's what I would do.
The coefficients {{{5}}} and {{{-21}}} (at the ends) multiply to give {{{5*(-21)=-105}}}.
I will look for pairs of factors of {{{-105}}} that add up to {{{-8}}} , which is the middle coefficient.
{{{105=1*105=3*35=5*21=7*15}}}
Since {{{7*(15)=-105}}} and {{{7+(-18)=-8}}} , {{{7}}} and {{{-15}}} will be the coefficients that I need.
I factor {{{5x^2-8x-21}}} "by parts" as follows: 
{{{5x^2-8x-21=5x^2-15x+7x-21=(5x^2-15x)+(7x-21)=5x(x-3)+7(x-3)=(5x+7)(x-3)}}}
Then I re-write the equation {{{5x^2-8x-21=0}}} as
{{{(5x+7)(x-3)=0}}} and realize that for the product to be zero, one factor must be zero,
so either {{{x-3=0}}}-->{{{highlight(x=3)}}} ,
or {{{5x+7=0}}}-->{{{5x=-7}}}-->{{{highlight(x=-7/5)}}}
 
COMPLETING THE SQUARE sometimes is very easy, it always works, and does not require remembering formulas.
It is not too easy in this case either, but here it goes:
{{{5x^2-8x-21=0}}}-->{{{5x^2-8x=21}}}-->{{{(5x^2-8x)/5=21/5}}}-->{{{x^2-(8/5)x=21/5}}}
{{{x^2-(8/5)x}} is part of the perfect square {{{x^2-(8/5)x+(4/5)^2=(x-4/5)^2}}}
so, from {{{x^2-(8/5)x=21/5}}} ,
adding {{{(4/5)^2=16/25}}} to both sides of the equal sign, I get the equivalent equation
{{{x^2-(8/5)x+(4/5)^2=21/5+16/25}}} , which I can re-write as
{{{(x-4/5)^2=105/25+16/25}}}-->{{{(x-4/5)^2=121/25}}}-->{{{(x-4/5)^2=(11/5)^2}}}
So either {{{x-4/5=11/5}}}-->{{{x=4/5+11/5}}}-->{{{x=15/5}}}-->{{{highlight(x=3)}}} ,
or {{{x-4/5=-11/5}}}-->{{{x=4/5-11/5}}}-->{{{highlight(x=-7/5)}}}
 
THE QUADRATIC FORMULA says that the solutions to an equation of the form
{{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
That is easy enough if you
remember the formula,
have a calculator handy,
and have the knowledge and ability to use it without making mistakes.
In this case, we have {{{5x^2-8x-21=0}}} ,
so {{{a=5}}}, {{{b=-8}}} and {{{c=-21}}} .
{{{x = (-(-8) +- sqrt((-8)^2-4*5*(-21) ))/(2*5)=(8 +- sqrt(64+420 ))/10=(8 +- sqrt(484))/10=(8 +- 22)/10}}}
The solutions are
{{{x=(8+22)/10=30/10=highlight(3)}}}
and {{{x=(8-22)/10=-14/10=highlight(-7/5)}}}