Solve for $x$: $\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.

Hint: We are given the equation $\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$. First subtract $\dfrac{1}{2x}$from both sides and simplify.

Complete step-by-step answer:
We are given $\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.
So now subtracting $\dfrac{1}{2x}$from both sides we get,
$\Rightarrow$ $\dfrac{1}{2a+b+2x}-\dfrac{1}{2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}-\dfrac{1}{2x}$
Simplifying we get,
$\Rightarrow$$\dfrac{1}{2a+b+2x}-\dfrac{1}{2x}=\dfrac{1}{2a}+\dfrac{1}{b}$
Now taking LCM and solving we get,
$\Rightarrow$$\dfrac{2x-2a-b-2x}{4ax+2bx+4{{x}^{2}}}=\dfrac{2a+b}{2ab}$
Simplifying we get,
$\Rightarrow$$\dfrac{-\left( 2a+b \right)}{4ax+2bx+4{{x}^{2}}}=\dfrac{2a+b}{2ab}$
Divide whole both sides by $2a+b$and simplifying we get,
$\Rightarrow$$\dfrac{-1}{4ax+2bx+4{{x}^{2}}}=\dfrac{1}{2ab}$
Now cross multiplying we get,
$\Rightarrow$$4ax+2bx+4{{x}^{2}}=-2ab$
Now adding both sides by $2ab$ we get,
$\Rightarrow$$4ax+2bx+4{{x}^{2}}+2ab=-2ab+2ab$
Simplifying we get,
$\Rightarrow$$4ax+2bx+4{{x}^{2}}+2ab=0$
Now taking and splitting the equation we get,
$\Rightarrow$$4x\left( a+x \right)+2b\left( x+a \right)=0$
Now taking common we get,
$\Rightarrow$$(4x+2b)(x+a)=0$
So, we get,
$(4x+2b)=0$ and $(x+a)=0$
Now taking we get,
$x=\dfrac{-b}{2}$ and $x=-a$
Therefore, we get the value of $x$ as $\dfrac{-b}{2}$ and $-a$.

information:
Quadratic Formula helps to evaluate the solution of quadratic equations replacing the factorization method. A quadratic equation is of the form of $a{{x}^{2}}+bx+c=0$, where $a,b$ and $c$ are real numbers, also called “numeric coefficients”. We know that a second-degree polynomial will have at most two zeros. Therefore, a quadratic equation will have at most two roots. By splitting the middle term, we can factorize quadratic polynomials.
Note:

Note: The term ${{b}^{2}}-4ac$ in the quadratic formula is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation reveals the nature of roots.
If the value of discriminant $=0$ i.e. ${{b}^{2}}-4ac=0$ the quadratic equation will have equal roots.
If the value of discriminant $<0$ i.e. ${{b}^{2}}-4ac<0$ the quadratic equation will have imaginary roots.
If the value of discriminant $>0$ i.e. ${{b}^{2}}-4ac>0$ then the quadratic equation will have real roots.