Answered By Alireza R

The quadratic equation can be re-arranged as follows:

x (-x/90+4/3)=0

Above equation has two roots, one is x=0.

To find the other root, this equation should be resolved: -x/90+4/3=0

solvong above equation gives x=120.

So one root is 0 and one is 120

Answered By Charles S

Hello! First off, thank you for asking!  The other answer is incorrect because the question asks to solve by completing the square. First step I would highly recommend when solving a problem like this is to make sure that any coefficients (numbers in front of variables(letters)) that are fractions be moved in front of the variables to make it very clear. 

  $y=\frac{-1}{90}x^2+\frac{4}{3}x$y=−190 x²+43 x  

Because you specified that this equation should be solved by completing the square, the next step would be to factor the coefficients out.  If you ever have a choice, I would recommend using the other methods(graphing, factoring or the quadratic formula) to solve this because sometimes completing the square can be confusing.

 $y=-\frac{1}{90}\left(x^2-120x\right)$y=−190 (x²−120x) $x^2-120x$x²−120x   This step is called factoring. I factored the -1/90 from both terms.  I got -120 because 4/3 divided by -1/90 is -120.

The next step is to choose a number that will make  $x^2-120x$x²−120x a perfect square trinomial.  An easy way to figure this out is to take 120 and divide by 2 and then square it.  So 120 divide by 2 is 60.  60 squared is 3600.  Add and subtract 3600 inside the brackets.  Remeber that when you add and subtract the same number it will give you zero.  Like if you had 4 sandwiches and somebody took 4 sandwiches from you you wouldn't have any left.  The fancy math way of saying this is called additive inverses. It's used a lot because it doesn't actually change  the equation, only modifies it to look a little bit nicer for this technique (completing the square).

 $-\frac{1}{90}\left(x^2-120x+3600-3600\right)$−190 (x²−120x+3600−3600) 

Now comes the tricky part.  You have to remove the -3600 from the brackets and in the process multiply it by the -1/90. You have to multiply it by the -1/90 because of another math principle called the distributive property. It's basically a way that you multiply a monomial (one term) by a polynomial(many terms). So, you have this:

  $y=-\frac{1}{90}\left(x^2-120x+3600\right)+-\frac{1}{90}\left(-3600\right)$y=−190 (x²−120x+3600)+−190 (−3600)  

  $y=-\frac{1}{90}\left(x^2-120x+3600\right)+40$y=−190 (x²−120x+360‍0)+40  

We're so close! Now you have to factor that perfect square trinomial you made earlier.

 $y=-\frac{1}{90}\left(x-60\right)^2+40$y=‍−190 (x−60)²+40 

Remember when you learned about factoring?  (x^2-120x+3600) = (x-60)(x-60) = (x-60)^2

Now, your question specifically says to solve.  I'm going to assume that what your question meant to say was to solve for the roots or maybe solve for the x intercepts.  This means that you should set y=0 because x intercepts are on a graph when y=0. Remember (x,0) would be a point when the y=0 on a graph and it would hit the x axis?

  $-40=-\frac{1}{90}\left(x-60\right)^2$−40=−190 (x−60)²  

Now comes the fun part. You just need to rearrange the equation for x.  You learned how to do this earlier when using something called the square root principle. Basically, this principle told you to square root both sides to get rid of a squared on one side of the equation. 

  $-40=-\frac{1}{90}\left(x-60\right)^2$−40=−190 (x−60)² 

 $3600=\left(x-60\right)^2$3600=(x−60)^2‍  I multiplied both sides by -90 to get this and eliminate the fraction on the right. (mutiplicative inverses)

 $\sqrt{3600}=\sqrt{\left(x-60\right)^2}$√3600=√(x−60)²  square root principle. Just square root both sides to get rid of the squared on the right.

 $\pm60=x-60$±60=x−60 so close! add 60 to both sides

 $\pm60+60=x$±60+60=x 😁

 $x=120,x=0$x=120,x=0   Careful and don't rush this step.  It means +60+60 and -60+60.  

Hope this helps!