Ok so the first thing you should do is take de therm 2x-5 and pass it to the other side of the equation multiplying. After that you can distribute the product like this:
[tex]3=x\cdot(2\cdot x-5)=2x^2-5x[/tex]
Then you can pass the 3 to the other side so you have a square function equal to 0:
[tex]2x^2-5x-3=0[/tex]
This equation means that you have to look for the roots of a square function. Given a general square function:
[tex]ax^2+bx+c[/tex]
Its roots (also named zeroes) are given by:
[tex]x_1,x_2=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a}[/tex]
In our problem we have a=2, b=-5 and c=-3. Therefore the roots of the square funtion are:
[tex]x_1,x_2=\frac{-(-5)\pm\sqrt[]{(-5)^2-4\cdot2\cdot(-3)}}{2\cdot2}=\frac{5\pm\sqrt[]{25+24}}{4}=\frac{5\pm7}{4}[/tex]
The symbol between 5 and 7 means that one root is calculate by adding and the other by substracting:
[tex]x_1=\frac{5+7}{4}=3[/tex][tex]x_2=\frac{5-7}{4}=-0.5[/tex]
So the roots are 3 and -0.5 and are the two possible values for x. Therefore the correct answer is item A.
