# Solve the equation -4y+133+-42x+-2y^2+3x^2=0

## Answer

$y_1=\frac{4+ \sqrt{336x+16}}{-4}\:\:,\:\:y_2=\frac{4- \sqrt{336x+16}}{-4}$

## Step-by-step explanation

Problem

$3x^2-2y^2-42x-4y+133=0$

1

Subtract $133$ from both sides of the equation

$-4y-42x-2y^2+3x^2=-133$

## Answer

$y_1=\frac{4+ \sqrt{336x+16}}{-4}\:\:,\:\:y_2=\frac{4- \sqrt{336x+16}}{-4}$