Answer
$\{1,2\}$.
Work Step by Step
Substitute $2x-5 = A$ in the given equation.
$\Rightarrow A^2+4A+3=0$
Rewrite the middle term $4A$ as $3A+1A$.
$\Rightarrow A^2+3A+A+3=0$
Group terms.
$\Rightarrow (A^2+3A)+(A+3)=0$
Factor each term.
$\Rightarrow A(A+3)+1(A+3)=0$
Factor out $(A+3)$.
$\Rightarrow (A+3)(A+1)=0$
Set each factor equal to zero.
$\Rightarrow A+3=0$ or $A+1=0$
Isolate $A$.
$\Rightarrow A=-3$ or $A=-1$
Substitute back $A=2x-5$.
$\Rightarrow 2x-5=-3$ or $2x-5=-1$
Isolate $x$.
$\Rightarrow x=\frac{-3+5}{2}$ or $x=\frac{-1+5}{2}$
Simplify.
$\Rightarrow x=\frac{2}{2}$ or $x=\frac{4}{2}$
$\Rightarrow x=1$ or $x=2$.
The solution set is $\{1,2\}$.
