7/2t+4=5+3/4t

Solution for 7/2t+4=5+3/4t equation:

7/2t+4=5+3/4t

We move all terms to the left:

7/2t+4-(5+3/4t)=0


Domain of the equation: 2t!=0

t!=0/2

t!=0

t∈R



Domain of the equation: 4t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

7/2t-(3/4t+5)+4=0

We get rid of parentheses

7/2t-3/4t-5+4=0

We calculate fractions


28t/8t^2+(-6t)/8t^2-5+4=0

We add all the numbers together, and all the variables

28t/8t^2+(-6t)/8t^2-1=0

We multiply all the terms by the denominator


28t+(-6t)-1*8t^2=0

Wy multiply elements

-8t^2+28t+(-6t)=0

We get rid of parentheses

-8t^2+28t-6t=0

We add all the numbers together, and all the variables

-8t^2+22t=0

a = -8; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-8)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-8}=\frac{-44}{-16}
=2+3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-8}=\frac{0}{-16}
=0
$