J(x)=2x2+4x+1

Solution for J(x)=2x2+4x+1 equation:

(J)=2J^2+4J+1

We move all terms to the left:

(J)-(2J^2+4J+1)=0

We get rid of parentheses

-2J^2+J-4J-1=0

We add all the numbers together, and all the variables

-2J^2-3J-1=0

a = -2; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·(-2)·(-1)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-2}=\frac{2}{-4}
=-1/2
$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-2}=\frac{4}{-4}
=-1
$