2f(-8f-15)=3(4f-5)

Solution for 2f(-8f-15)=3(4f-5) equation:

2f(-8f-15)=3(4f-5)

We move all terms to the left:

2f(-8f-15)-(3(4f-5))=0

We multiply parentheses

-16f^2-30f-(3(4f-5))=0


We calculate terms in parentheses: -(3(4f-5)), so:

3(4f-5)

We multiply parentheses

12f-15

Back to the equation:

-(12f-15)


We get rid of parentheses

-16f^2-30f-12f+15=0

We add all the numbers together, and all the variables

-16f^2-42f+15=0

a = -16; b = -42; c = +15;
Δ = b2-4ac
Δ = -422-4·(-16)·15
Δ = 2724
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:

$\sqrt{\Delta}=\sqrt{2724}=\sqrt{4*681}=\sqrt{4}*\sqrt{681}=2\sqrt{681}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{681}}{2*-16}=\frac{42-2\sqrt{681}}{-32}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{681}}{2*-16}=\frac{42+2\sqrt{681}}{-32}
$