(2t-3/t+5)=0

Solution for (2t-3/t+5)=0 equation:

(2t-3/t+5)=0


Domain of the equation: t+5)!=0

t∈R


We get rid of parentheses

2t-3/t+5=0

We multiply all the terms by the denominator


2t*t+5*t-3=0

We add all the numbers together, and all the variables

5t+2t*t-3=0

Wy multiply elements

2t^2+5t-3=0

a = 2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·2·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*2}=\frac{-12}{4}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*2}=\frac{2}{4}
=1/2
$