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-10=0

We multiply all the terms by the denominator


-10*3*f+20=0

Wy multiply elements

-30f*f+20=0

Wy multiply elements

-30f^2+20=0

a = -30; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-30)·20
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*-30}=\frac{0-20\sqrt{6}}{-60}
=-\frac{20\sqrt{6}}{-60}
=-\frac{\sqrt{6}}{-3}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*-30}=\frac{0+20\sqrt{6}}{-60}
=\frac{20\sqrt{6}}{-60}
=\frac{\sqrt{6}}{-3}
$