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

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Answer 1 · 2018-05-10T15:03:19

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

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

Divide by 4

$x^2+5x-y^2+3/2y=3/4$

Complete the square

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

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

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

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