2f(5f-2)-10(f-3f+6)=-8f(f+4)+4(2f-7f)

Solution for 2f(5f-2)-10(f-3f+6)=-8f(f+4)+4(2f-7f) equation:

2f(5f-2)-10(f-3f+6)=-8f(f+4)+4(2f-7f)

We move all terms to the left:

2f(5f-2)-10(f-3f+6)-(-8f(f+4)+4(2f-7f))=0

We add all the numbers together, and all the variables

2f(5f-2)-10(-2f+6)-(-8f(f+4)+4(-5f))=0

We multiply parentheses

10f^2-4f+20f-(-8f(f+4)+4(-5f))-60=0


We calculate terms in parentheses: -(-8f(f+4)+4(-5f)), so:

-8f(f+4)+4(-5f)

We multiply parentheses

-8f^2-32f-20f

We add all the numbers together, and all the variables

-8f^2-52f

Back to the equation:

-(-8f^2-52f)


We add all the numbers together, and all the variables

10f^2-(-8f^2-52f)+16f-60=0

We get rid of parentheses

10f^2+8f^2+52f+16f-60=0

We add all the numbers together, and all the variables

18f^2+68f-60=0

a = 18; b = 68; c = -60;
Δ = b2-4ac
Δ = 682-4·18·(-60)
Δ = 8944
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:

$\sqrt{\Delta}=\sqrt{8944}=\sqrt{16*559}=\sqrt{16}*\sqrt{559}=4\sqrt{559}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4\sqrt{559}}{2*18}=\frac{-68-4\sqrt{559}}{36}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4\sqrt{559}}{2*18}=\frac{-68+4\sqrt{559}}{36}
$