F(x)=2x-10+x(5x-7)

Solution for F(x)=2x-10+x(5x-7) equation:

(F)=2F-10+F(5F-7)

We move all terms to the left:

(F)-(2F-10+F(5F-7))=0


We calculate terms in parentheses: -(2F-10+F(5F-7)), so:

2F-10+F(5F-7)

determiningTheFunctionDomain
2F+F(5F-7)-10

We multiply parentheses

5F^2+2F-7F-10

We add all the numbers together, and all the variables

5F^2-5F-10

Back to the equation:

-(5F^2-5F-10)


We get rid of parentheses

-5F^2+F+5F+10=0

We add all the numbers together, and all the variables

-5F^2+6F+10=0

a = -5; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-5)·10
Δ = 236
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:

$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{59}}{2*-5}=\frac{-6-2\sqrt{59}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{59}}{2*-5}=\frac{-6+2\sqrt{59}}{-10}
$