Question 452589
Let the numbers be {{{j}}} and {{{k}}}
given:
(1) {{{ j + k = 3 }}}
(2) {{{ j*k = -180 }}}
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(1) {{{ j + k = 3 }}}
(1) {{{ k = 3 - j }}}
Substitute this into (2)
(2) {{{ j*( 3-j ) = -180 }}}
(2) {{{ 3j - j^2 = -180 }}}
(2) {{{ -j^2 + 3j + 180 = 0 }}}
Solve using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = -1 }}}
{{{ b = 3 }}}
{{{ c = 180 }}}
{{{x = (-3 +- sqrt( 3^2-4*(-1)*180 ))/(2*(-1)) }}}
{{{x = (-3 +- sqrt( 9 + 720 ))/(-2) }}}
{{{x = (-3 +- sqrt( 729 ))/(-2) }}}
{{{ j = ( -3 - 27 ) / (-2) }}}
{{{ j = (-30)/(-2) }}}
{{{ j = 15 }}}
and, since 
{{{ j + k = 3 }}}
{{{ k = 3 - 15 }}}
{{{ k = -12 }}}
check the answer:
(2) {{{ j*k = -180 }}}
(2) {{{ 15*(-12) = -180 }}}
(2) {{{ -180 = -180 }}}
OK
The numbers are 15 and -12