Calculate $\sqrt{2x-3}$ to the power of $2$ and get $2x-3$.
$$2x-3=\left(3-2x\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3-2x\right)^{2}$.
$$2x-3=9-12x+4x^{2}$$
Subtract $9$ from both sides.
$$2x-3-9=-12x+4x^{2}$$
Subtract $9$ from $-3$ to get $-12$.
$$2x-12=-12x+4x^{2}$$
Add $12x$ to both sides.
$$2x-12+12x=4x^{2}$$
Combine $2x$ and $12x$ to get $14x$.
$$14x-12=4x^{2}$$
Subtract $4x^{2}$ from both sides.
$$14x-12-4x^{2}=0$$
Divide both sides by $2$.
$$7x-6-2x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-2x^{2}+7x-6=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx-6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=7$$ $$ab=-2\left(-6\right)=12$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $12$.
$$1,12$$ $$2,6$$ $$3,4$$
Calculate the sum for each pair.
$$1+12=13$$ $$2+6=8$$ $$3+4=7$$
The solution is the pair that gives sum $7$.
$$a=4$$ $$b=3$$
Rewrite $-2x^{2}+7x-6$ as $\left(-2x^{2}+4x\right)+\left(3x-6\right)$.
$$\left(-2x^{2}+4x\right)+\left(3x-6\right)$$
Factor out $2x$ in the first and $-3$ in the second group.
$$2x\left(-x+2\right)-3\left(-x+2\right)$$
Factor out common term $-x+2$ by using distributive property.
$$\left(-x+2\right)\left(2x-3\right)$$
To find equation solutions, solve $-x+2=0$ and $2x-3=0$.
$$x=2$$ $$x=\frac{3}{2}$$
Substitute $2$ for $x$ in the equation $\sqrt{2x-3}=3-2x$.
$$\sqrt{2\times 2-3}=3-2\times 2$$
Simplify. The value $x=2$ does not satisfy the equation because the left and the right hand side have opposite signs.
$$1=-1$$
Substitute $\frac{3}{2}$ for $x$ in the equation $\sqrt{2x-3}=3-2x$.
