# How do you solve 5x^2+3x-4=0 by completing the square?

The roots are

${x}_{1} = \frac{- 3 + \sqrt{89}}{10}$
${x}_{2} = \frac{- 3 - \sqrt{89}}{10}$

#### Explanation:

From the given
$5 {x}^{2} + 3 x - 4 = 0$

factor out 5 from the first two terms

$5 \left({x}^{2} + \frac{3}{5} x\right) - 4 = 0$

Use now the number 3/5. Divide this number by 2 then square the result to obtain 9/100. This 9/100 will be added and subtracted to terms inside the grouping symbol.

Let us continue

$5 \left({x}^{2} + \frac{3}{5} x\right) - 4 = 0$

$5 \left({x}^{2} + \frac{3}{5} x + \frac{9}{100} - \frac{9}{100}\right) - 4 = 0$

We now have a perfect square trinomial

${x}^{2} + \frac{3}{5} x + \frac{9}{100} = {\left(x + \frac{3}{10}\right)}^{2}$ so that

$5 \left({\left(x + \frac{3}{10}\right)}^{2} - \frac{9}{100}\right) - 4 = 0$

Transpose now the -4 to the right side then divide by 5 both sides of the equation then transpose the -9/100 to the right also.

$5 \left({\left(x + \frac{3}{10}\right)}^{2} - \frac{9}{100}\right) = 4$

$\frac{\cancel{5} \left({\left(x + \frac{3}{10}\right)}^{2} - \frac{9}{100}\right)}{\cancel{5}} = \frac{4}{5}$

${\left(x + \frac{3}{10}\right)}^{2} - \frac{9}{100} = \frac{4}{5}$

${\left(x + \frac{3}{10}\right)}^{2} = \frac{4}{5} + \frac{9}{100}$

simplify

${\left(x + \frac{3}{10}\right)}^{2} = \frac{80 + 9}{100}$

${\left(x + \frac{3}{10}\right)}^{2} = \frac{89}{100}$
Extract the square root of both sides

$\sqrt{{\left(x + \frac{3}{10}\right)}^{2}} = \pm \sqrt{\frac{89}{100}}$

$x + \frac{3}{10} = \pm \sqrt{\frac{89}{100}}$

$x = - \frac{3}{10} \pm \frac{\sqrt{89}}{10}$

The roots are

${x}_{1} = - \frac{3}{10} + \frac{\sqrt{89}}{10}$
${x}_{2} = - \frac{3}{10} - \frac{\sqrt{89}}{10}$

God bless....I hope the explanation is useful.