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?

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.
