Question 7 (1 point) Solve: \( -3(x+8)^{2}=-108_{\text {by square roots. }} \) \( \mathrm{x}= \) \( \qquad \) and \( \qquad \) Blank 1: \( \square \) Blank 2: \( \square \) Question 8 (1 point) Solve the equation by factoring. \[ c^{2}-4 c=0 \] a \( c=1 \) or \( c=-\sqrt{4} \) b \( c=0 \) or \( c=-4 \) c \( c=0 \) or \( c=4 \) d \( c=0 \) or \( c=\sqrt{4} \)

To solve the first equation \( -3(x+8)^{2}=-108 \), first divide both sides by -3 to get \( (x+8)^{2}=36 \). Taking the square root, we find \( x+8=6 \) or \( x+8=-6 \). This leads to the solutions \( x=6-8=-2 \) and \( x=-6-8=-14 \). Now let’s dive into the second equation \( c^{2}-4c=0 \). Factoring it, we rewrite it as \( c(c-4)=0 \). This gives us two solutions: either \( c=0 \) or \( c-4=0 \) leading to \( c=4 \). So the correct answer here would be option **c: \( c=0 \) or \( c=4 \)**.