Question 1143812
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With vertex (h,k), the vertex form of the equation is<br>
{{{(x-h) = (1/(4p))(y-k)^2}}}<br>
Put the equation in that form by completing the square in y and then solving for x.<br>
{{{y^2+3x-2y+7 = 0}}}  given
{{{y^2-2y = -3x-7}}}  isolate the y terms
{{{y^2-2y+1 = -3x-7+1 = -3x-6}}}  complete the square in y
{{{(y-1)^2 = -3(x+2)}}}  put in vertex form
{{{x+2 = (-1/3)(y-1)^2}}}  done....<br>
The vertex is (h,k) = (-2,1).