Question 1038879
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Solve by two methods, quadratic equation and factoring:
2x^2-x-15=0
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As I just said in <A HREF=https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1038843.html>https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1038843.html</A>, 

solve on your own at least by the quadratic formula.


Demonstrate that you can do this arithmetic.


<U>Comment from student</U>:

<pre>
Solve by 2 methods,Quad Equation and Factoring : 2x^2-x-15=0 
You ask me to show, this may or may not be right 

X=-b±&#8730;b^2-4ac/2a. a=2, b= -1, c= -15 X=-(-1)±&#8730;1^2+4^2(-15)/2^2 X1,2=(1±11)/2^2 X1=12/4= 3 X2=-10/4= -5/2 

Having problems with getting my zero Factoring right and checking problem 
</pre>


<U>My responce</U>: Bravo that you found a courage to start it.


Your calculations are incorrect, and I see that you really need help. Let me fix it:


<pre>
X = {{{(1 +- sqrt((-1)^2 - 4*2*(-15)))/4}}} = {{{(1 +- sqrt(1 + 120))/4}}} = {{{(1 +- sqrt(121))/4}}} = {{{(1 +- 11)/4}}}.

{{{x[1]}}} = {{{12/4}}} = {{{3}}},  {{{x[2]}}} = {{{-10/4}}} = {{{-5/2}}}.
</pre>

<pre>
Now factoring:

{{{2x^2-x-15}}} = {{{2*(x-x[1])*(x-x[2])}}} = {{{2*(x - 3)*(x-(-5/2))}}} = {{{(x-3)*(2x+5)}}}.
</pre>