F(x)=2x^2+16x

Solution for F(x)=2x^2+16x equation:

(F)=2F^2+16F

We move all terms to the left:

(F)-(2F^2+16F)=0

We get rid of parentheses

-2F^2+F-16F=0

We add all the numbers together, and all the variables

-2F^2-15F=0

a = -2; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·(-2)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*-2}=\frac{0}{-4}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*-2}=\frac{30}{-4}
=-7+1/2
$