Question

What is the real # solution

Answer 1

`\bf (x+5)(x+6)=0\implies \begin{cases} x+5=0\implies &x=-5\\ x+6=0\implies &x=-6 \end{cases} \\\\\\ (x-9)(x+8)=0\implies \begin{cases} x-9=0\implies &x=9\\ x+8=0\implies &x=-8 \end{cases} \\\\\\ (x+4)(x-4)=0\implies \begin{cases} x+4=0\implies &x=-4\\ x-4=0\implies &x=4 \end{cases} \\\\\\ x^2-14x+49\implies (x-7)^2\iff (x-7)(x-7) \\\\\\ (x-7)(x-7)=0\implies \begin{cases} x-7=0\implies &x=7\\ x-7=0\implies &x=7 \end{cases}`