20-2t=1/2t+28;t

Solution for 20-2t=1/2t+28;t equation:

20-2t=1/2t+28t

We move all terms to the left:

20-2t-(1/2t+28t)=0


Domain of the equation: 2t+28t)!=0

t∈R


We add all the numbers together, and all the variables

-2t-(+28t+1/2t)+20=0

We get rid of parentheses

-2t-28t-1/2t+20=0

We multiply all the terms by the denominator


-2t*2t-28t*2t+20*2t-1=0

Wy multiply elements

-4t^2-56t^2+40t-1=0

We add all the numbers together, and all the variables

-60t^2+40t-1=0

a = -60; b = 40; c = -1;
Δ = b2-4ac
Δ = 402-4·(-60)·(-1)
Δ = 1360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{85}}{2*-60}=\frac{-40-4\sqrt{85}}{-120}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{85}}{2*-60}=\frac{-40+4\sqrt{85}}{-120}
$