Question 1177943
{{{y = 2x^2 -5x - 6}}} ...........eq.1
{{{y = 3x+ 4}}}........................eq.2
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{{{y = 2x^2 -5x - 6}}} ...........eq.1...........substitute {{{y}}} from eq.2

{{{3x+ 4= 2x^2 -5x - 6}}}

{{{0= 2x^2 -5x-3x - 6-4}}}

{{{2x^2 -8x - 10=0}}} .......factor

{{{2(x^2 -4x - 5)=0}}} .........factor

{{{2 (x - 5) (x + 1) = 0}}}

so, solutions are {{{x=5}}} and {{{x=-1}}}

then

{{{y = 3*5+ 4=19}}}
{{{y = 3*(-1)+ 4=1 }}}

points of intersection: ({{{5}}},{{{19}}}) and ({{{-1}}},{{{1}}})

the length is equal to distance between ({{{5}}},{{{19}}}) and ({{{-1}}},{{{1}}})

use distance formula

{{{d=sqrt((-1-5)^2+(1-19)^2)}}}

{{{d=sqrt((-6)^2+(-18)^2)}}}

{{{d=sqrt(360)}}}

{{{d=sqrt(36*10)}}}

{{{d=6sqrt(10)}}}=>the length of the line segment formed by the points of intersection for the system