Question 691174
{{{2e^x + 4xe^x + (x^2-1)e^x = 0}}}....first multiply {{{(x^2-1)e^x}}}

{{{2e^x + 4xe^x + x^2e^x-1e^x= 0}}}

 {{{x^2e^x + 4xe^x +e^x= 0}}}

{{{e^x*x^2 + 4e^x*x +e^x= 0}}}....here you have a quadratic equation (in general form {{{ax^2+bx+c=0}}}), where {{{a=e^x}}}, {{{b=4e^x}}} and {{{c=e^x}}}

now you use quadratic formula to solve for {{{x}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ....plug in values for {{{a}}}, {{{b}}} and {{{c}}}


{{{x = (-4e^x +- sqrt((4e^x)^2-4*e^x*e^x ))/(2*e^x) }}}


{{{x = (-4e^x +- sqrt(16e^(2x)-4*e^(2x) ))/(2e^x) }}}


{{{x = (-4e^x +- sqrt(12e^(2x) ))/(2e^x) }}}....since {{{e^(2x)=(e^x)^2}}}, we will have


{{{x = (-4e^x +- sqrt(4*3(e^x)^2 ))/(2e^x) }}}


{{{x = (-4e^x +- 2(e^x)sqrt(3 ))/(2e^x) }}}.......factor out {{{(1/2)}}}


{{{x = (1/2)((-4cross((e^x)) +- 2cross((e^x))sqrt(3 ))/cross((e^x))) }}}.....cancel {{{(e^x)}}}


{{{x = (1/2)(-4 +- 2sqrt(3 )) }}}

solutions:

{{{x = (1/2)(-4 + 2sqrt(3 )) }}}


{{{x = (1/2)(-4 + 2*1.73) }}}


{{{x = (1/2)(-4 + 3.46) }}}


{{{x = (1/2)(-0.54) }}}

{{{x =-0.27}}}............if you plug in {{{x =-0.27}}} in given equation, you will get answer {{{-0.00541999}}} which is rounded to whole number equal to zero

or

{{{x = (1/2)(-4 -2sqrt(3 )) }}}


{{{x = (1/2)(-4-2*1.73) }}}


{{{x = (1/2)(-4 - 3.46) }}}


{{{x = (1/2)(-7.46) }}}


{{{x =-3.73}}}...........if you plug in {{{x =-3.73}}}, you get {{{-0.000170349}}}, you also round it to zero

the difference appears because rounding decimals during calculation of {{{x}}}