Question 1100221
Let {{{ a }}} = 1st number 
Let {{{ b }}} = 2nd number
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(1) {{{ 2a^2 - b = -3 }}}
(2) {{{ ( a + 5 )^2 = b - 2 }}}
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(1) {{{ b = 2a^2 + 3 }}}
Plug (1) into (2)
(2) {{{ ( a + 5 )^2 = 2a^2 + 3 - 2 }}}
(2) {{{ a^2 + 10a + 25 = 2a^2 + 1 }}}
(2) {{{ a^2 - 10a - 24 = 0 }}}
(2) {{{ ( a - 12 )*( a + 2 ) = 0 }}} ( by looking at it )
{{{ a = 12 }}}
{{{ a = -2 }}}
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(1) {{{ 2*(-2)^2 - b = -3 }}}
(1) {{{ 2*4 - b = -3 }}}
(1) {{{ 8 - b = -3 }}}
(1) {{{ b = 11 }}}
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also:
(1) {{{ 2*(12)^2 - b = -3 }}}
(1) {{{ 288 - b = -3 }}}
(1) {{{ b = 291 }}}
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check:
(2) {{{ ( a+5 )^2 = b - 2 }}}
(2) {{{ ( -2+5 )^2 = 11 - 2 }}}
(2) {{{ 3^2 = 11 - 2 }}}
(2) {{{ 9 = 9 }}}
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(2) {{{ ( 12+ 5 )^2 = 291 - 2 }}}
(2) {{{ 289 = 291 - 2 }}}
(2) {{{ 289 = 289 }}}
OK
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There are 2 pairs of solutions:
{{{ a = -2 }}}
{{{ b = 11 }}}
and
{{{ a = 12 }}}
{{{ b = 291 }}}
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check the math & get another opinion if needed