2t+8+(1/2t)-t=11

Solution for 2t+8+(1/2t)-t=11 equation:

2t+8+(1/2t)-t=11

We move all terms to the left:

2t+8+(1/2t)-t-(11)=0


Domain of the equation: 2t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

2t+(+1/2t)-t+8-11=0

We add all the numbers together, and all the variables

t+(+1/2t)-3=0

We get rid of parentheses

t+1/2t-3=0

We multiply all the terms by the denominator


t*2t-3*2t+1=0

Wy multiply elements

2t^2-6t+1=0

a = 2; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·2·1
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{7}}{2*2}=\frac{6-2\sqrt{7}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{7}}{2*2}=\frac{6+2\sqrt{7}}{4}
$