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?

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 $x=\frac 4 9$.
