# Solve the equation (x+1)(x-2)=0

## \left(x+1\right)\left(x-2\right)=0

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$x_1=2,\:x_2=-1$

## Step by step solution

Problem

$\left(x+1\right)\left(x-2\right)=0$
1

Multiplying polynomials $x$ and $x+1$

$-2-2x+x+x^2=0$
2

Subtract $-2$ from both sides of the equation

$-2x+x+x^2=2$
3

Adding $-2x$ and $x$

$x^2-x=2$
4

Rewrite the equation

$-2-x+x^2=0$
5

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

$x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
6

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

$x=\frac{-1\left(-1\right)\pm \sqrt{8+{\left(-1\right)}^2}}{2}$
7

Multiply $-1$ times $-1$

$x=\frac{1\pm \sqrt{8+{\left(-1\right)}^2}}{2}$
8

Calculate the power

$x=\frac{1\pm \sqrt{8+1}}{2}$
9

Add the values $1$ and $8$

$x=\frac{1\pm \sqrt{9}}{2}$
10

Calculate the power

$x=\frac{1\pm 3}{2}$
11

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

$x_1=\frac{1+ 3}{2}\:\:,\:\:x_2=\frac{1- 3}{2}$
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Simplifying

$x_1=2,\:x_2=-1$
13

We found that the two real solutions of the equation are

$x_1=2,\:x_2=-1$

$x_1=2,\:x_2=-1$

Polynomials

0.26 seconds

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