The sum of the squares of two negative numbers is 106 and the difference of the squares of the number is 56 find the numbers

Calculation:
Let two negative numbers are x and y.
The sum of the square of two negative numbers is 106.
$x^2+y^2=106$ (1)
And the difference of the square of the number is 56.
$x^2-y^2=56$ (2)
Add equation (1) and (2).
$(x^2+y^2)+(x^2-y^2)=106+56$
$2x^2+0=162$
$2x^2=162$
$(2x^2)/2=162/2$ [Divide by 2]
$x^2=81$
Taking square root.
$\sqrt{x^2}=\pm \sqrt{81}$
$x=\pm \sqrt{9^2}$
$x=\pm 9$
x=-9, because x is negative
Substitute x = -9 in equation (1).
$-9^2+y^2=106$
$81+y^2=106$
$-81+81+y^2=106-81$ [Substract 81 from both sides]
$=y^2=25$
Taking square root.
$\sqrt{y^2}=\pm \sqrt{25}$
$y=\sqrt{5^2}$
$y=\pm 5$
y=-5, because y is negative
Thus, the negative numbers are x = -9 and y = -5.