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

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

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

We move all terms to the left:

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


Domain of the equation: 7t!=0

t!=0/7

t!=0

t∈R



Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R


We add all the numbers together, and all the variables

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

We calculate fractions


20t/28t^2+21t/28t^2-7=0

We multiply all the terms by the denominator


20t+21t-7*28t^2=0

We add all the numbers together, and all the variables

41t-7*28t^2=0

Wy multiply elements

-196t^2+41t=0

a = -196; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·(-196)·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*-196}=\frac{-82}{-392}
=41/196
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*-196}=\frac{0}{-392}
=0
$