(2x*1)-(4x)2=(-3x)2+2x+1

Solution for (2x*1)-(4x)2=(-3x)2+2x+1 equation:

(2x*1)-(4x)2=(-3x)2+2x+1

We move all terms to the left:

(2x*1)-(4x)2-((-3x)2+2x+1)=0

We add all the numbers together, and all the variables

(+2x*1)-4x2-((-3x)2+2x+1)=0

We add all the numbers together, and all the variables

-4x^2+(+2x*1)-((-3x)2+2x+1)=0

We get rid of parentheses

-4x^2+2x*1-((-3x)2+2x+1)=0


We calculate terms in parentheses: -((-3x)2+2x+1), so:

(-3x)2+2x+1

We add all the numbers together, and all the variables

2x+(-3x)2+1

We multiply parentheses

2x-6x+1

We add all the numbers together, and all the variables

-4x+1

Back to the equation:

-(-4x+1)


Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2+6x-1=0

a = -4; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·(-4)·(-1)
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*-4}=\frac{-6-2\sqrt{5}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*-4}=\frac{-6+2\sqrt{5}}{-8}
$