\( (2x+3)(2x-3)=7 \)
\( x=\pm1 \)
\( x=\pm2 \)
Solve for x:\( (x+1)(2x+1)=2x^2+4 \)
\( x=1 \)
\( x=0 \)
\( (x+1)(x+3)-x=x^2 \)
\( x=-1 \)
(2x+3)(2x−3)=7 (2x+3)(2x-3)=7 (2x+3)(2x−3)=7
x=±2 x=\pm2 x=±2
Solve for x:(x+1)(2x+1)=2x2+4 (x+1)(2x+1)=2x^2+4 (x+1)(2x+1)=2x2+4
x=1 x=1 x=1
(x+1)(x+3)−x=x2 (x+1)(x+3)-x=x^2 (x+1)(x+3)−x=x2
x=−1 x=-1 x=−1