I did not choose the suitable pairs of binomials, as Math hunter did , so my solution is not short . Anyway 😊 (x+7)(x+8)(x+9)(x+10)=(x-7)(x-8)(x-9)(x-10) (x ²+15x+56)(x ²+19x+90)=(x ²-15x+56)(x ²-19x+90) (1) Let x ²+56=t and x ²+90=ω (1) => (t+15x)(ω+19x)=(t-15x)(ω-19x) …………………. x=0 or ω=-19t /15 Hence x ²+56=t and x ²+90=-19t /15 If you eliminate t between the equations , you will find x²=√(-71) => x=± i √71

And my son's solution , who is not afraid mathematical operations !!!! (x+7)(x+8)(x+9)(x+10)=(x-7)(x-8)(x-9)(x-10) => ................................................................ 68x³+4828x=0 68x(x²+71)=0 x=0 or x=±ⅈ√71