1/4t+1/5t-4=5

Solution for 1/4t+1/5t-4=5 equation:

1/4t+1/5t-4=5

We move all terms to the left:

1/4t+1/5t-4-(5)=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R



Domain of the equation: 5t!=0

t!=0/5

t!=0

t∈R


We add all the numbers together, and all the variables

1/4t+1/5t-9=0

We calculate fractions


5t/20t^2+4t/20t^2-9=0

We multiply all the terms by the denominator


5t+4t-9*20t^2=0

We add all the numbers together, and all the variables

9t-9*20t^2=0

Wy multiply elements

-180t^2+9t=0

a = -180; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-180)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-180}=\frac{-18}{-360}
=1/20
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-180}=\frac{0}{-360}
=0
$