(12/t)+t-8=0

Solution for (12/t)+t-8=0 equation:

(12/t)+t-8=0


Domain of the equation: t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+12/t)+t-8=0

We add all the numbers together, and all the variables

t+(+12/t)-8=0

We get rid of parentheses

t+12/t-8=0

We multiply all the terms by the denominator


t*t-8*t+12=0

We add all the numbers together, and all the variables

-8t+t*t+12=0

Wy multiply elements

t^2-8t+12=0

a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$