8t-3/t+1=0

Solution for 8t-3/t+1=0 equation:

8t-3/t+1=0


Domain of the equation: t!=0

t∈R


We multiply all the terms by the denominator


8t*t+1*t-3=0

We add all the numbers together, and all the variables

t+8t*t-3=0

Wy multiply elements

8t^2+t-3=0

a = 8; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·8·(-3)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{97}}{2*8}=\frac{-1-\sqrt{97}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{97}}{2*8}=\frac{-1+\sqrt{97}}{16}
$