Solution - Solving quadratic equations by completing the square

Use the standard form of a quadratic equation, $$a{x}^2+bx+c=0$$ , to find the coefficients:

$$5x^2-54=0$$

$$a=5$$
$$b=0$$
$$c=-54$$

Because $$a=5$$, divide all coefficients and constants on both sides of the equation by $$5$$:

$$5x^2+0x-54=0$$

$$\frac{5}{5}{x}^2+0\frac{x}{5}-\frac{54}{5}=\frac{0}{5}$$

$$x^2+0x-\frac{54}{5}=0$$

The coefficients are:
$$a=1$$
$$b=0$$
$$c=-\frac{54}{5}$$

Add $$\frac{54}{5}$$ to both sides of the equation:

$$x^2+0x-\frac{54}{5}=0$$

$$x^2+0x-\frac{54}{5}+\frac{54}{5}=0+\frac{54}{5}$$

$$x^2+0x=\frac{54}{5}$$

To make the left side of the equation into a perfect square trinomial, add a new constant equal to $${\left(\frac{b}{2}\right)}^2$$ to the equation:

$$b=0$$

Add $$0$$ to both sides of the equation:

$$x^2+0x=\frac{54}{5}$$

$$x^2+0x+0=\frac{54}{5}+0$$

Now we have perfect square trinomial, we can write it as a perfect square form by adding half of the $$b$$ coefficient, $$\frac{b}{2}$$ :
$$b=0$$

$$x^2+0x+0=\frac{54}{5}$$

$${\left(x+0\right)}^2=\frac{54}{5}$$

Take the square root of both sides of the equation: IMPORTANT: When finding the square root of a constant, we get two solutions: positive and negative

$${\left(x+0\right)}^2=\frac{54}{5}$$

$$\sqrt{{\left(x+0\right)}^2}=\sqrt{\frac{54}{5}}$$

$$x+0=\pm \sqrt{\frac{54}{5}}$$

$$x+0+0=\pm \sqrt{\frac{54}{5}}$$

$$x=\pm \sqrt{\frac{54}{5}}$$

$$x=0\pm \frac{\sqrt{54}}{\sqrt{5}}$$

$$x=0\pm \frac{\sqrt{54}·\sqrt{5}}{\sqrt{5}·\sqrt{5}}$$

$$x=0\pm \frac{\sqrt{270}}{5}$$

$$x=0\pm \frac{\sqrt{2·3·3·3·5}}{5}$$

$$x=0\pm \frac{\sqrt{2·3^2·3·5}}{5}$$

$$x=0\pm \frac{3·\sqrt{2·3·5}}{5}$$

$$x=0\pm \frac{3·\sqrt{6·5}}{5}$$

$$x=0\pm \frac{3·\sqrt{30}}{5}$$

$$x_1=0+\frac{3·\sqrt{30}}{5}$$
$$x_2=0-\frac{3·\sqrt{30}}{5}$$