# How do you write 4x^2 + 20x - 4y^2 + 6y - 3 = 0 in standard form?

May 10, 2018

${\left(x + \frac{5}{2}\right)}^{2} - {\left(y - \frac{3}{4}\right)}^{2} = \frac{121}{16}$

#### Explanation:

$4 {x}^{2} + 20 x - 4 {y}^{2} + 6 y = 3$

Divide by 4

${x}^{2} + 5 x - {y}^{2} + \frac{3}{2} y = \frac{3}{4}$

Complete the square

${\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} - {\left(y - \frac{3}{4}\right)}^{2} - \frac{9}{16} = \frac{3}{4}$

${\left(x + \frac{5}{2}\right)}^{2} - {\left(y - \frac{3}{4}\right)}^{2} = \frac{3}{4} + \frac{109}{16}$

${\left(x + \frac{5}{2}\right)}^{2} - {\left(y - \frac{3}{4}\right)}^{2} = \frac{121}{16}$