Solve the system of equations. $$ y=50 x^{2}-18 x-36 $$ $$ y=-18 x+14 $$ Write the coordinates in exact form. Simplify all fractions and radica ( ,

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Solve the system of equations $$ y=50x^{2}-18x-36;y=-18x+14 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}y=50x^{2}-18x-36\y=-18x+14\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$50x^{2}-18x-36=-18x+14$$
- step2: Cancel equal terms:
  $$50x^{2}-36=14$$
- step3: Move the constant to the right side:
  $$50x^{2}=14+36$$
- step4: Add the numbers:
  $$50x^{2}=50$$
- step5: Divide both sides:
  $$\frac{50x^{2}}{50}=\frac{50}{50}$$
- step6: Divide the numbers:
  $$x^{2}=1$$
- step7: Simplify the expression:
  $$x=\pm \sqrt{1}$$
- step8: Simplify:
  $$x=\pm 1$$
- step9: Separate into possible cases:
  $$x=1\cup x=-1$$
- step10: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=1\y=50x^{2}-18x-36\end{array}ight.\cup \left\{ \begin{array}{l}x=-1\y=50x^{2}-18x-36\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=-4\end{array}ight.\cup \left\{ \begin{array}{l}x=-1\y=32\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-1\y=32\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=-4\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-1\y=32\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=-4\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(-1,32ight)\cup \left(x,yight) = \left(1,-4ight)$$
The solutions to the system of equations are $$ (x,y) = (-1,32) $$ and $$ (x,y) = (1,-4) $$.

Therefore, the coordinates in exact form are $$ (-1,32) $$ and $$ (1,-4) $$.
The solutions are $$ (-1,32) $$ and $$ (1,-4) $$.

Solve the system of equations.

Write the coordinates in exact form. Simplify all fractions and radica
( ,

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Solve the system of equations. Write the coordinates in exact form. Simplify all fractions and radica ( ,