5f+10=(1/2)(20f-40)

Solution for 5f+10=(1/2)(20f-40) equation:

5f+10=(1/2)(20f-40)

We move all terms to the left:

5f+10-((1/2)(20f-40))=0


Domain of the equation: 2)(20f-40))!=0

f∈R


We add all the numbers together, and all the variables

5f-((+1/2)(20f-40))+10=0

We multiply parentheses ..

-((+20f^2+1/2*-40))+5f+10=0

We multiply all the terms by the denominator


-((+20f^2+1+5f*2*-40))+10*2*-40))=0


We calculate terms in parentheses: -((+20f^2+1+5f*2*-40)), so:

(+20f^2+1+5f*2*-40)

We get rid of parentheses

20f^2+5f*2*+1-40

We add all the numbers together, and all the variables

20f^2+5f*2*-39

Wy multiply elements

20f^2+10f^2-39

We add all the numbers together, and all the variables

30f^2-39

Back to the equation:

-(30f^2-39)


We add all the numbers together, and all the variables

-(30f^2-39)=0

We get rid of parentheses

-30f^2+39=0

a = -30; b = 0; c = +39;
Δ = b2-4ac
Δ = 02-4·(-30)·39
Δ = 4680
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:

$\sqrt{\Delta}=\sqrt{4680}=\sqrt{36*130}=\sqrt{36}*\sqrt{130}=6\sqrt{130}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{130}}{2*-30}=\frac{0-6\sqrt{130}}{-60}
=-\frac{6\sqrt{130}}{-60}
=-\frac{\sqrt{130}}{-10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{130}}{2*-30}=\frac{0+6\sqrt{130}}{-60}
=\frac{6\sqrt{130}}{-60}
=\frac{\sqrt{130}}{-10}
$