F(x)=(-2x+10)(-2x+15)

Solution for F(x)=(-2x+10)(-2x+15) equation:

(F)=(-2F+10)(-2F+15)

We move all terms to the left:

(F)-((-2F+10)(-2F+15))=0

We multiply parentheses ..

-((+4F^2-30F-20F+150))+F=0


We calculate terms in parentheses: -((+4F^2-30F-20F+150)), so:

(+4F^2-30F-20F+150)

We get rid of parentheses

4F^2-30F-20F+150

We add all the numbers together, and all the variables

4F^2-50F+150

Back to the equation:

-(4F^2-50F+150)


We add all the numbers together, and all the variables

F-(4F^2-50F+150)=0

We get rid of parentheses

-4F^2+F+50F-150=0

We add all the numbers together, and all the variables

-4F^2+51F-150=0

a = -4; b = 51; c = -150;
Δ = b2-4ac
Δ = 512-4·(-4)·(-150)
Δ = 201
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{-(51)-\sqrt{201}}{2*-4}=\frac{-51-\sqrt{201}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+\sqrt{201}}{2*-4}=\frac{-51+\sqrt{201}}{-8}
$