6=t+4/5t-3

Solution for 6=t+4/5t-3 equation:

6=t+4/5t-3

We move all terms to the left:

6-(t+4/5t-3)=0


Domain of the equation: 5t-3)!=0

t∈R


We get rid of parentheses

-t-4/5t+3+6=0

We multiply all the terms by the denominator


-t*5t+3*5t+6*5t-4=0

Wy multiply elements

-5t^2+15t+30t-4=0

We add all the numbers together, and all the variables

-5t^2+45t-4=0

a = -5; b = 45; c = -4;
Δ = b2-4ac
Δ = 452-4·(-5)·(-4)
Δ = 1945
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{-(45)-\sqrt{1945}}{2*-5}=\frac{-45-\sqrt{1945}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{1945}}{2*-5}=\frac{-45+\sqrt{1945}}{-10}
$