How to solve (2x-3)^(2)=18

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (2x-3)^(2)=18 | Equatix

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$${ \left(2x-3 \right) }^{ 2 } =18$$

$x = \frac{3 \sqrt{2} + 3}{2} \approx 3.621320344$
$x=\frac{3-3\sqrt{2}}{2}\approx -0.621320344$

Take the square root of both sides of the equation.

$$2x-3=3\sqrt{2}$$ $$2x-3=-3\sqrt{2}$$

Add $3$ to both sides of the equation.

$$2x-3-\left(-3\right)=3\sqrt{2}-\left(-3\right)$$ $$2x-3-\left(-3\right)=-3\sqrt{2}-\left(-3\right)$$

Subtracting $-3$ from itself leaves $0$.

$$2x=3\sqrt{2}-\left(-3\right)$$ $$2x=-3\sqrt{2}-\left(-3\right)$$

Subtract $-3$ from $3\sqrt{2}$.

$$2x=3\sqrt{2}+3$$

Subtract $-3$ from $-3\sqrt{2}$.

$$2x=3-3\sqrt{2}$$

Divide both sides by $2$.

$$\frac{2x}{2}=\frac{3\sqrt{2}+3}{2}$$ $$\frac{2x}{2}=\frac{3-3\sqrt{2}}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x=\frac{3\sqrt{2}+3}{2}$$ $$x=\frac{3-3\sqrt{2}}{2}$$