# Solve the equation y-1y^2=y^2-3

## y-y^2=y^2-3

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$y_1=-1,\:y_2=1.5$

## Step by step solution

Problem

$y-y^2=y^2-3$
1

Rewrite the equation

$-\left(y^2-3\right)+y-y^2=0$
2

Multiply $\left(y^2+-3\right)$ by $-1$

$y-y^2+3-y^2=0$
3

Adding $-y^2$ and $-y^2$

$y+3-2y^2=0$
4

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=-2$, $b=1$ and $c=3$

$y =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
5

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

$y=\frac{-1\pm \sqrt{3\left(-2\right)\left(-4\right)+1^2}}{-2\cdot 2}$
6

Multiply $-6$ times $-4$

$y=\frac{-1\pm \sqrt{24+1^2}}{-4}$
7

Calculate the power

$y=\frac{-1\pm \sqrt{24+1}}{-4}$
8

Add the values $1$ and $24$

$y=\frac{-1\pm \sqrt{25}}{-4}$
9

Calculate the power

$y=\frac{-1\pm 5}{-4}$
10

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

$y_1=\frac{-1+ 5}{-4}\:\:,\:\:y_2=\frac{-1- 5}{-4}$
11

Simplifying

$y_1=-1,\:y_2=1.5$
12

We found that the two real solutions of the equation are

$y_1=-1,\:y_2=1.5$

$y_1=-1,\:y_2=1.5$

Polynomials

0.35 seconds

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