What is (2x^2+7x)+(2y^2+4y)=6 ?

The solution to (2x^2+7x)+(2y^2+4y)=6 is Circle with (a,b)=(-7/4 ,-1),r=(sqrt(113))/4

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(2x^2+7x)+(2y^2+4y)=6

(2 x^{2} + 7 x) + (2 y^{2} + 4 y) = 6

(a, b) = (- \frac{7}{4}, - 1), r = \frac{113}{4}

Solution steps

2 x^{2} + 7 x + (2 y^{2} + 4 y) = 6

Rewrite

2 x^{2} + 7 x + (2 y^{2} + 4 y) = 6

in the form of the standard circle equation

(x - (- \frac{7}{4}))^{2} + (y - (- 1))^{2} = (\frac{113}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{4}, - 1), r = \frac{113}{4}

(x - 1)^{2} + (y + 5)^{2} = 4^{2}

(Y - 1)^{2} + (X + 5)^{2} = 1

(x + 2)^{2} + (y - 1)^{2} = 2^{2}

0 + y^{2} + z^{2} = 1

(x - 2)^{2} + (y - 17)^{2} = 64

What is (2x^2+7x)+(2y^2+4y)=6 ?
The solution to (2x^2+7x)+(2y^2+4y)=6 is Circle with (a,b)=(-7/4 ,-1),r=(sqrt(113))/4