Solve the equation 8t-9+t^2=0

t^2+8t-9=0

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Answer

$t_1=1,\:t_2=-9$

Step by step solution

Problem

$t^2+8t-9=0$
1

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=1$, $b=8$ and $c=-9$

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

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

$t=\frac{8\left(-1\right)\pm \sqrt{36+8^2}}{2}$
3

Multiply $-1$ times $8$

$t=\frac{-8\pm \sqrt{36+8^2}}{2}$
4

Calculate the power

$t=\frac{-8\pm \sqrt{36+64}}{2}$
5

Add the values $64$ and $36$

$t=\frac{-8\pm \sqrt{100}}{2}$
6

Calculate the power

$t=\frac{-8\pm 10}{2}$
7

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

$t_1=\frac{-8+ 10}{2}\:\:,\:\:t_2=\frac{-8- 10}{2}$
8

Simplifying

$t_1=1,\:t_2=-9$
9

We found that the two real solutions of the equation are

$t_1=1,\:t_2=-9$

Answer

$t_1=1,\:t_2=-9$

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Problem Analysis

Main topic:

Quadratic formula

Time to solve it:

0.24 seconds

Views:

123