F(x)=2x^2+(3x)^2

Solution for F(x)=2x^2+(3x)^2 equation:

(F)=2F^2+(3F)^2

We move all terms to the left:

(F)-(2F^2+(3F)^2)=0

determiningTheFunctionDomain
-(2F^2+3F^2)+F=0

We get rid of parentheses

-2F^2-3F^2+F=0

We add all the numbers together, and all the variables

-5F^2+F=0

a = -5; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-5)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-5}=\frac{-2}{-10}
=1/5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-5}=\frac{0}{-10}
=0
$