Question

radical din x la 2 +x-4=2radical din 2 :) help

Answer 1

$\sqrt{x^2+x-4} =2 \sqrt{2} \\ ( \sqrt{x^2+x-4} )^2=(2 \sqrt{2} )^2 \\ x^2+x-4=8 \\ x^2+x=8+4 \\ x^2+x=12 \\ x^2+x-12=0 \\ (x+4)(x-3)=0 \\ x+4=0 \\ x=0-4 \\ x=-4 \\ x-3=0 \\ x=0+3 \\ x=3 \\ x=-4;x=3$