已知关于X的方程X²+2(m+2)X+m²—5=0有两个实数根,并且两个根的平方和比这两个根的积大1

let x1,x2 be roots of eqution x²+2(m+2)x+m²-5=0
x1+x2 = -2(m+2)
x1x2 = m^2-5
两个根的平方和比这两个根的积大16
x1^2 + x2^2 - x1x2 = 16
(x1+x2)^2 -3x1x2 =16
4(m+2)^2 - 3(m^2-5) =16
m^2+16m+15=0
(m-1)(m-15)=0
m=1 or 15