Solve: $3 y^{2} + 14 y + 8 = 0$

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Solution

$3 y^{2} + 14 y + 8$

$= 3 y^{2} + 14 y + 8$

$= 3 y^{2} + 12 y + 2 y + 8$ (splitting $14 y = 12 y + 2 y$ )

$= 3 y (y + 4) + 2 (y + 4)$

$= (y + 4) (3 y + 2)$

for zero of the quadratic function
$3 y^{2} + 14 y + 8 = 0$

so $(y + 4) (3 y + 2) = 0$

$y + 4 = 0$ or $3 y + 2 = 0$

$y = - 4$ $y = - \frac{2}{3}$

$y = - 4, - \frac{2}{3}$

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Factorise:

$3 y^{2} + 14 y + 8$

3 y^{2} - 14 y + 8 = 0

14 y - 8 = 13

The point of intersection of the lines represented by the equation $2 x^{2} + 3 y^{2} + 7 x y + 8 x + 14 y + 8 = 0$ is

4 + y - 14 y^{2} = 0

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