# How do you complete the square to solve 0=5x^2 + 2x - 3?

Jul 3, 2015

$x = \frac{3}{5}$ or $x = - 1$

#### Explanation:

Step 1. Write your equation in standard form.

$5 {x}^{2} + 2 x - 3 = 0$

Step 2. Move the constant to the right hand side of the equation.

Add $3$ to each side .

$5 {x}^{2} + 2 x - 3 + 3 = 0 + 3$

$5 {x}^{2} + 2 x = 3$

Step 3. Divide both sides of the equation by the coefficient of ${x}^{2}$.

Divide both sides by 5.

${x}^{2} + \frac{2}{5} x = \frac{3}{5}$

Step 4. Square the coefficient of x and divide by 4.

${\left(\frac{2}{5}\right)}^{2} / 4 = \frac{\frac{4}{25}}{4} = \frac{1}{25}$

Step 5. Add the result to each side.

${x}^{2} + \frac{2}{5} x + \frac{1}{25} = \frac{3}{5} + \frac{1}{25}$

${x}^{2} + \frac{2}{5} x + \frac{1}{25} = \frac{15}{25} + \frac{1}{25}$

${x}^{2} + \frac{2}{5} x + \frac{1}{25} = \frac{16}{25}$

Step 6. Take the square root of each side.

x+1/5 = ±4/5

Case 1

${x}_{1} + \frac{1}{5} = + \frac{4}{5}$

${x}_{1} = \frac{4}{5} - \frac{1}{5} = \frac{4 - 1}{5}$

${x}_{1} = \frac{3}{5}$

Case 2

${x}_{2} + \frac{1}{5} = - \frac{4}{5}$

${x}_{2} = - \frac{4}{5} - \frac{1}{5} = \frac{- 4 - 1}{5} = \frac{- 5}{5}$

${x}_{2} = - 1$

So $x = \frac{3}{5}$ or $x = - 1$

Check: Substitute the values of $x$ back into the quadratic.

(a) $x = \frac{3}{5}$

$5 {x}^{2} + 2 x - 3 = 5 {\left(\frac{3}{5}\right)}^{2} + 2 \left(\frac{3}{5}\right) - 3 = 5 \left(\frac{9}{25}\right) + \frac{6}{5} - 3 = \frac{9}{5} + \frac{6}{5} - \frac{15}{5} = \frac{9 + 6 - 15}{5} = 0$.

(b) $x = - 1$

5x^2 + 2x -3 = 5(-1)^2 + 2(-1) -3 = 5(1) – 2 -3 = 5-2-3 = 0