Find the real solutions to the equations below using any method of your choice.

1.  $2x^2=2−3x$ 

   Step 1(a):

   Rewrite the equation in standard form.
   $2x^2+3x=2−3x+3x$
   $2x^2+3x=2$
   $2x^2+3x−2=2−2$
   $2x^2+3x−2=0$

   Solve by Zero Product Property:

   Step 2 (Zero Product Property):

   Factor the equation:
   $2x^2+3x−2=0$
   $(2x−1)(x+2)=0$

   Step 3 (Zero Product Property):

   Get the roots from the factors.
   $(2x−1)=0$
   $x=0.5$
   
   $(x+2)=0$
   $x=−2$

   Solve by Quadratic Formula:

   Step 4 (Quadratic Formula):

   In the quadratic formula, substitute the coefficient of $x^2$ for $a$, the coefficient of $x$ for $b$, and the third term for $c$.

   $x = \frac {-b \pm \sqrt {b^2 - 4 ac}}{2a}$

   $x = \frac {-3\; \pm\; \sqrt {3^2\; -\; 4(2)(-2)}}{2(2)}$

   Step 5 (Quadratic Formula):

   Simplify.

   $\textit{x}\; =\; \frac {-3\; \pm\; \sqrt {9\; +\; 16}}{4}$

   $x = \frac {-3\; \pm\; \sqrt{25}}{4}$

   $x = \frac {-3\; \pm\; 5}{4}$

   $x=\left\{\frac{2}{4},-\frac{8}{4}\right\}$

   Answer (a):

   $\textit{x}=\frac{1}{2}, -2$

2.  $(2x−3)^2+4=0$ 

   Hint (b):

   See the help for part (a).

   More Help:

   This quadratic is not factorable. What happens when you use the other two methods? Are there any real solutions?