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

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Answer 1 · 2018-05-10T15:03:19

${(x + \frac{5}{2})}^{2} - {(y - \frac{3}{4})}^{2} = \frac{121}{16}$ $(x+5/2)^2-(y-3/4)^2=[121]/[16]$

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

Divide by 4

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

Complete the square

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

${(x + \frac{5}{2})}^{2} - {(y - \frac{3}{4})}^{2} = \frac{3}{4} + \frac{109}{16}$ $(x+5/2)^2-(y-3/4)^2=3/4+[109]/[16]$

${(x + \frac{5}{2})}^{2} - {(y - \frac{3}{4})}^{2} = \frac{121}{16}$ $(x+5/2)^2-(y-3/4)^2=[121]/[16]$

How do you write 4x2+20x−4y2+6y−3=04x^2 + 20x - 4y^2 + 6y - 3 = 0 in standard form?