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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+8F^2-20F+6F-15))+4=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 get rid of parentheses

-8F^2+14F+15+4=0

We add all the numbers together, and all the variables

-8F^2+14F+19=0

a = -8; b = 14; c = +19;
Δ = b2-4ac
Δ = 142-4·(-8)·19
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{201}}{2*-8}=\frac{-14-2\sqrt{201}}{-16}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{201}}{2*-8}=\frac{-14+2\sqrt{201}}{-16}
$