((t^2+6t+2)-2)/t-2=12

Solution for ((t^2+6t+2)-2)/t-2=12 equation:

((t^2+6t+2)-2)/t-2=12

We move all terms to the left:

((t^2+6t+2)-2)/t-2-(12)=0


Domain of the equation: t!=0

t∈R


We add all the numbers together, and all the variables

((t^2+6t+2)-2)/t-14=0

We multiply all the terms by the denominator


((t^2+6t+2)-2)-14*t=0


We calculate terms in parentheses: +((t^2+6t+2)-2), so:

(t^2+6t+2)-2

We get rid of parentheses

t^2+6t+2-2

We add all the numbers together, and all the variables

t^2+6t

Back to the equation:

+(t^2+6t)


We add all the numbers together, and all the variables

-14t+(t^2+6t)=0

We get rid of parentheses

t^2-14t+6t=0

We add all the numbers together, and all the variables

t^2-8t=0

a = 1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*1}=\frac{0}{2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*1}=\frac{16}{2}
=8
$