Question 1132331
Try changing one of the variables so that is clear than  it is not equal to 0.  You do not want to say  1 is not equal to zero.


-(L+m)-Lx^2+(2L+m)x=0


{{{-(L+m)-Lx^2+(2L+m)x=0}}}


{{{-Lx^2+(2L+m)x-(L+m)=0}}}


{{{Lx^2-(2L+m)x+(L+m)=0}}}

Just apply the general solution formula to this.


for variable x in quadratic equation {{{ax^2+bx+c=0}}}, solutions for x are

{{{x=(-b+- sqrt(b^2-4ac))/(2a)}}}.


Your example will give this according to this formula,
{{{x=(-2L-m+- sqrt((2L+m)^2-4L(L+m)))/(2L)}}}

Simplify this as much as possible.

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What can you do with the radical expression?

{{{sqrt((2L+m)^2-4L(L+m))}}}

{{{sqrt(4L^2+4mL+m^2-4L^2-4mL)}}}

{{{sqrt(4L^2-4L^2+4mL-4mL+m^2)}}}

You should have no trouble from here.  You continue and finish!