Solve for X and Y

I have the following equation:

$$0 = 34x^2+92xy+68y^2−250x−344y+461$$

I cannot find any way to get the values of both x and y from this equation, any help would be much appreciated, especially a step by step solution.

Edit: I'm quite new to this site, if there's any way for me to improve this question please edit it or let me know!

• One way to improve the post would be to present your work. Also, what do we know about $x$ and $y$? Complex, real, integer...? This equation surely has a continuum of complex roots. You can view it as a quadratic equation in $x$ and solve in terms of $y$ as a parameter. – A. Pongrácz May 2 at 16:07
• They are all real numbers. Would a quadratic equation work with 10 or more variables? – Ruler Of The World May 2 at 16:09
• You are changing your post in the wrong direction. Without showing any of your own efforts, you should definitely not ask for a step-by-step solution. As they say in shark tank, "I'm out". – A. Pongrácz May 2 at 16:11
• Of course, my apologies. I'll keep the question the same and open another one if necessary. Let's focus on just two variables – Ruler Of The World May 2 at 16:14

Solving for $$x$$ with the usual formula,

$$x=\frac{-(92y-250)\pm\sqrt{(92y-250)^2-4\cdot34\cdot(68y^2-344y+461)}}{68}.$$

After simplification, the discriminant is

$$-196(2y-1)^2,$$ which is only non-negative when $$y=\frac12.$$

$$x$$ follows.

• Great answer! What is the formula you used to do this? – Ruler Of The World May 2 at 16:20
• @RulerOfTheWorld: as said in the comments. – Yves Daoust May 2 at 16:21
• As far as I know, the quadratic formula is: x = (-b ± √(b²-4ac))/2a, however what you have used doesn't seem to be the same – Ruler Of The World May 2 at 16:26
• @RulerOfTheWorld: it is exactly that one. What differences do you see ? – Yves Daoust May 2 at 17:33
• Sorry for the late reply. I'm not sure where you got the values of a b and c from for the formula, I thought a=34 (from 34x^2), b=-250 (from -250x) and c=461 (from +461) – Ruler Of The World May 2 at 18:24

$$Q^T D Q = H$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 23 }{ 17 } & 1 & 0 \\ - \frac{ 125 }{ 34 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 34 & 0 & 0 \\ 0 & \frac{ 98 }{ 17 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 23 }{ 17 } & - \frac{ 125 }{ 34 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 34 & 46 & - 125 \\ 46 & 68 & - 172 \\ - 125 & - 172 & 461 \\ \end{array} \right)$$

So $$x + \frac{ 23 }{ 17 } y - \frac{ 125 }{ 34 } = 0$$ $$y - \frac{ 1 }{ 2 } = 0$$ because 34 times the square of the first one plus $$\frac{ 98 }{ 17 }$$ times the square of the second must be zero.

The relationship between the matrices $$D,H$$ is called congruence