Exit Ticket Solve each equation. 1. \( (x-2)^{2}=400 \) 2. \( 3 x^{2}+1=25 \) 3. \( (q+7)(q-4)=0 \) 4. \( z^{2}-3 z=18 \)

To solve the first equation \( (x-2)^{2}=400 \), take the square root of both sides. This gives you two cases: \( x - 2 = 20 \) and \( x - 2 = -20 \). Solving these, we find \( x = 22 \) and \( x = -18 \). For the second equation \( 3x^{2}+1=25 \), first subtract 1 from both sides to get \( 3x^{2} = 24 \). Then, divide by 3 to find \( x^{2} = 8 \). Taking the square root, we get two solutions: \( x = 2\sqrt{2} \) and \( x = -2\sqrt{2} \). In the third equation \( (q+7)(q-4)=0 \), use the zero product property which states that if a product equals zero, at least one of the factors must equal zero. This gives us \( q + 7 = 0 \) leading to \( q = -7 \), and \( q - 4 = 0 \) leading to \( q = 4 \). Lastly, for the equation \( z^{2}-3z=18 \), first move everything to one side to get \( z^{2}-3z-18=0 \). We can factor this into \( (z-6)(z+3)=0 \), giving us \( z = 6 \) and \( z = -3 \). So, the solutions are: 1. \( x = 22, -18 \) 2. \( x = 2\sqrt{2}, -2\sqrt{2} \) 3. \( q = -7, 4 \) 4. \( z = 6, -3 \)