F(x)=(4x+3)(2x-5)

Solution for F(x)=(4x+3)(2x-5) equation:

(F)=(4F+3)(2F-5)

We move all terms to the left:

(F)-((4F+3)(2F-5))=0

We multiply parentheses ..

-((+8F^2-20F+6F-15))+F=0


We calculate terms in parentheses: -((+8F^2-20F+6F-15)), so:

(+8F^2-20F+6F-15)

We get rid of parentheses

8F^2-20F+6F-15

We add all the numbers together, and all the variables

8F^2-14F-15

Back to the equation:

-(8F^2-14F-15)


We add all the numbers together, and all the variables

F-(8F^2-14F-15)=0

We get rid of parentheses

-8F^2+F+14F+15=0

We add all the numbers together, and all the variables

-8F^2+15F+15=0

a = -8; b = 15; c = +15;
Δ = b2-4ac
Δ = 152-4·(-8)·15
Δ = 705
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{705}}{2*-8}=\frac{-15-\sqrt{705}}{-16}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{705}}{2*-8}=\frac{-15+\sqrt{705}}{-16}
$