Solve equation $\sqrt{4t + 1} = 3 - 3t$ Solve equation $\sqrt{4t + 1} = 3-3t$  
→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$
→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$
I'm stuck at that point. Can someone tell me what I am doing wrong?
 A: For starters, squaring $\sqrt{4t+1}=3-3t$ gives $4t+1=9-18t+9t^2$. A little rearranging and we have $9t^2-22t+8=0$.
With a bit of insight we can see that this is $(9x-4)(x-2)$.
However, I'm pretty crap at factoring so lets use the quadratic formula:
$$ax^2+bx+c=0$$ implies 
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
So we take $$x=\frac{-(-22)\pm\sqrt{22^2-4(9)(8)}}{2(9)}$$ now step-by-step we have
$$x=\frac{22\pm\sqrt{484-288}}{18}=\frac{22\pm\sqrt{196}}{18}=\frac{22\pm14}{18}
=\frac{11\pm7}{9}$$
So for "$+$" we have $x=\frac{18}{9}=2$ and for "$-$" we have $x=\frac 4 9$.
A: You made a mistake in squaring the right hand side:
$$
(3-3t)^2=9-18t+9t^2
$$
However you have also to impose the condition $3-3t\ge0$, because the left hand side is non negative by definition. So you get
$$
9t^2-22t+8=0,\qquad t\le1
$$
The quadratic equation has roots $2$ and $4/9$, as it's easy to see. Hence the unique solution to your equation is $t=4/9$.
If you "back substitute", you can understand what happens: for $t=2$ the left hand side is $\sqrt{4\cdot2+1}=3$, while the right hand side is $3-3\cdot2=-3$.
For $t=4/9$, the left hand side is $\sqrt{16/9+1}=5/3$, and the right hand side is $3-3\cdot(4/9)=3-4/3=5/3$.
The problem is that upon squaring you also add the solutions to $\sqrt{4t+1}=-(3-3t)$ that is another equation.
A: Hint:
If $ax^2+bx+c=0$, $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
A: Given: $\sqrt{4t+1}=3-3t$ 
$$\bigg(\sqrt{4t+1}\bigg)^2=\bigg(3-3t\bigg)^2$$
$$4t+1=\bigg(3-3t\bigg)\bigg(3-3t\bigg)$$
$$4t+1=9-9t-9t+9t^2$$
$$4t+1=9-18t+9t^2$$
$$0=9-1-18t-4t+9t^2$$
$$0=9t^2-22t+8$$
Which is a quadratic in the form:$$0 =ax^2+bx+c$$ 
Use the quadratic equation:
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ 
$$x = \frac{ -(-22)\pm \sqrt{(-22)^2-4(9)(8)}}{2(9)}$$
$$x = \frac{22\pm\sqrt{484-288}}{18}$$
$$x = \frac{22\pm\sqrt{196}}{18}$$
$$x = \frac{22\pm{14}}{18}$$
$$x = 2 $$ $$or$$ $$x =\frac{4}{9} $$
Understand that it still be checked that these are indeed solutions to your initial equation.
