(20/3)f-5=5

Solution for (20/3)f-5=5 equation:

(20/3)f-5=5

We move all terms to the left:

(20/3)f-5-(5)=0


Domain of the equation: 3)f!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

(+20/3)f-5-5=0

We add all the numbers together, and all the variables

(+20/3)f-10=0

We multiply parentheses

20f^2-10=0

a = 20; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·20·(-10)
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*20}=\frac{0-20\sqrt{2}}{40}
=-\frac{20\sqrt{2}}{40}
=-\frac{\sqrt{2}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*20}=\frac{0+20\sqrt{2}}{40}
=\frac{20\sqrt{2}}{40}
=\frac{\sqrt{2}}{2}
$