20-2t=1/2t+28

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

20-2t=1/2t+28

We move all terms to the left:

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


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

t∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-4t^2-16t-1=0

a = -4; b = -16; c = -1;
Δ = b2-4ac
Δ = -162-4·(-4)·(-1)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{15}}{2*-4}=\frac{16-4\sqrt{15}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{15}}{2*-4}=\frac{16+4\sqrt{15}}{-8}
$