Question 289118
Under square root(3-x) + under square root(x+5)=4

I think you mean the following:
{{{ sqrt(3-x) + sqrt(x+5) = 4 }}}
set a = 3 - x
set b = x + 5
{{{ sqrt(a) + sqrt(b) = 4 }}} (square both sides)
{{{ a + 2sqrt(a)sqrt(b) + b = 16 }}}
{{{ a + 2sqrt(ab) + b = 16 }}}
{{{ 2sqrt(ab) = 16 - a - b }}} (square both sides)
{{{ 4ab = (16 - a - b)^2 }}} (evaluate (16 - a - b)^2)
{{{ (16 - a - b)(16 - a - b) }}}
{{{ 256 - 16a - 16b - 16a + a^2 + ab - 16b + ab + b^2 }}}
{{{ a^2 + b^2 + 2ab - 32a - 32b + 256 }}} (returning)
{{{ 4ab = (16 - a - b)^2 }}}
{{{ 4ab = a^2 + b^2 + 2ab - 32a - 32b + 256 }}}
{{{ 0 = a^2 + b^2 - 2ab - 32a - 32b + 256 }}}
{{{ a^2 = (3 - x)^2 }}}
{{{ a^2 = 9 - 6x + x^2 }}}
{{{ a^2 = x^2 - 6x + 9 }}}
{{{ b^2 = (x + 5)^2 }}}
{{{ b^2 = x^2 + 10x + 25 }}}
{{{ ab = (3 - x)(x + 5) }}}
{{{ ab = 3x + 15 - x^2 - 5x }}}
{{{ ab = -x^2 - 2x + 15 }}}
{{{ 0 = (x^2 - 6x + 9) + (x^2 + 10x + 25) - 2(-x^2 - 2x + 15) - 32(-x + 3) - 32(x + 5) + 256 }}}
{{{ 0 = (x^2 - 6x + 9) + (x^2 + 10x + 25) + (2x^2 + 4x - 30) + (32x - 96) + (-32x - 160) + 256 }}}
{{{ 0 = 2x^2 + 4x + 34 + (2x^2 + 4x - 30) - 256 + 256 }}}
{{{ 0 = 4x^2 + 8x + 4 }}}
{{{ 0 = x^2 + 2x + 1}}}
{{{ 0 = (x + 1)(x + 1) }}}
x = -1
check: {{{ sqrt(3-x) + sqrt(x+5) = 4 }}}
{{{ sqrt(3 - -1) + sqrt(-1 +5) }}}
{{{ sqrt(4) + sqrt(4) }}}
{{{ 2 + 2 = 4 }}}