Solving for x with exponents (algebra) So I am trying to help a friend do her homework and I am a bit stuck. 
$$8x+3 = 3x^2$$
I can look at this and see that the answer is $3$, but I am having a hard time remembering how to solve for $x$ in this situation. 
Could someone be so kind as to break down the steps in solving for $x$.
Thanks in advance for replies. 

This is a quadratic equation: the highest power of the unknown is $2$. Rearrange it to bring everything to one side of the equation:
$$3x^2-8x-3=0\;.$$
If you can easily factor the resulting expression, you can take a shortcut, but otherwise you either complete the square or use the quadratic formula.
Completing the square relies on the fact that $(x+a)^2=x^2+2ax+a^2$. First factor out the coefficient of $x^2$:
$$3x^2-8x-3=3\left(x^2-\frac83x-1\right)\;.\tag{1}$$
Now notice that if you set $a=-\dfrac{8/3}2=-\dfrac43$, you’ll have $$(x+a)^2=\left(x-\frac43\right)^2=x^2-\frac83x+\frac{16}9\;,$$ which agrees in all but the constant term with the expression in parentheses in $(1)$. Thus,
$$x^2-\frac83x-1=\left(x-\frac43\right)^2-\frac{16}9-1=\left(x-\frac43\right)^2-\frac{25}9\;,$$
and on substituting back into $(1)$ we have
$$0=3x^2-8x-3=3\left(x-\frac43\right)^2-\frac{25}3\;.$$ Rearranging this gives us
$$3\left(x-\frac43\right)^2=\frac{25}3\;,$$ or $$\left(x-\frac43\right)^2=\frac{25}9\;.$$ Finally, taking the square root on both sides and remembering that there are two square roots, one positive and one negative, we get
$$x-\frac43=\pm\frac53$$ and therefore $$x=\frac53+\frac43=\frac93=3\quad\text{or}\quad x=-\frac53+\frac43=-\frac13\;.$$
The quadratic formula can be derived by applying the method of completing the square to the general quadratic equation $ax^2+bx+c=0$. The result is that
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;.$$
