# Solve the equation -4y^2-8y+150x+25x^2=-129

## 25x^2+150x-4y^2-8y=-129

Go!
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$y_1=-6.7663,\:y_2=4.7663$

## Step by step solution

Problem

$25x^2+150x-4y^2-8y=-129$
1

Rewrite the equation

$129-8y-4y^2=0$
2

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=-4$, $b=-8$ and $c=129$

$y =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
3

Substituting the values of the coefficients of the equation in the quadratic formula

$y=\frac{-8\left(-1\right)\pm \sqrt{129\left(-4\right)\left(-4\right)+{\left(-8\right)}^2}}{-4\cdot 2}$
4

Multiply $-1$ times $-8$

$y=\frac{8\pm \sqrt{2064+{\left(-8\right)}^2}}{-8}$
5

Calculate the power

$y=\frac{8\pm \sqrt{2064+64}}{-8}$
6

Add the values $64$ and $2064$

$y=\frac{8\pm \sqrt{2128}}{-8}$
7

Calculate the power

$y=\frac{8\pm 46.1303}{-8}$
8

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$y_1=\frac{8+ 46.1303}{-8}\:\:,\:\:y_2=\frac{8- 46.1303}{-8}$
9

Simplifying

$y_1=-6.7663,\:y_2=4.7663$
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We found that the two real solutions of the equation are

$y_1=-6.7663,\:y_2=4.7663$

$y_1=-6.7663,\:y_2=4.7663$