F(x)=2/5x+10

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

(F)=2/5F+10

We move all terms to the left:

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


Domain of the equation: 5F+10)!=0

F∈R


We get rid of parentheses

F-2/5F-10=0

We multiply all the terms by the denominator


F*5F-10*5F-2=0

Wy multiply elements

5F^2-50F-2=0

a = 5; b = -50; c = -2;
Δ = b2-4ac
Δ = -502-4·5·(-2)
Δ = 2540
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:

$\sqrt{\Delta}=\sqrt{2540}=\sqrt{4*635}=\sqrt{4}*\sqrt{635}=2\sqrt{635}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{635}}{2*5}=\frac{50-2\sqrt{635}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{635}}{2*5}=\frac{50+2\sqrt{635}}{10}
$