Solve the equation (x^2-3x)/2-5=(x-20)/4

\frac{x^2-3x}{2}-5=\frac{x-20}{4}

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Answer

$x_1=1.7913,\:x_2=-2.7913$

Step by step solution

Problem

$\frac{x^2-3x}{2}-5=\frac{x-20}{4}$
1

Grouping terms

$-\frac{x-20}{4}-5+\frac{x^2-3x}{2}=0$
2

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 $c=-5$

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

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

$x=\frac{-1\pm \sqrt{20+1^2}}{2}$
4

Calculate the power

$x=\frac{-1\pm \sqrt{20+1}}{2}$
5

Add the values $1$ and $20$

$x=\frac{-1\pm \sqrt{21}}{2}$
6

Calculate the power

$x=\frac{-1\pm \sqrt{21}}{2}$
7

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

$x_1=\frac{-1+ \sqrt{21}}{2}\:\:,\:\:x_2=\frac{-1- \sqrt{21}}{2}$
8

Simplifying

$x_1=1.7913,\:x_2=-2.7913$
9

We found that the two real solutions of the equation are

$x_1=1.7913,\:x_2=-2.7913$

Answer

$x_1=1.7913,\:x_2=-2.7913$

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Problem Analysis

Main topic:

Quadratic formula

Time to solve it:

0.24 seconds

Views:

91