Question 1020900
They are somewhat alike, and if you understand how to handle either, you can handle the other.


{{{abs((2x+(1/3))x+0.3)=3.5}}}


Distribute.
{{{abs(2x^2+x/3+0.3)=3.5}}}
Maybe easier if clear the fractions denominators...
{{{3*abs(2x^2+x/3+0.3)=3*(3.5)}}}
{{{abs(6x^2+x+0.9)=10.5}}}, not perfectly, but good enough to continue comfortably.


Either the input to the absolute value is non-negative, or it is negative.  Solve for each of these two cases.



NON-NEGATIVE
{{{6x^2+x+0.9=10.5}}}
{{{6x^2+x+0.9-10.5=0}}}
{{{6x^2+x-9.6=0}}}
Discrim is {{{1+4*6*9.6=231.4}}}


NEGATIVE
{{{-6x^2-x-0.9=10.5}}}
{{{6x^2+x+0.9=-10.5}}}
{{{6x^2+x+0.9+10.5=0}}}
{{{6x^2+x+11.4=0}}}
Discrim is {{{1-4*6*11.4=negativeValueDiscriminant}}}
These will not be real numbers.


Continue for the non-negative inputs to the absolute value.
{{{highlight(x=(-1+- sqrt(231.4))/12)}}}


You may want to check both values in case one of them does not work in the original equation.