x(x+3)+1=3(x+x^2)

Solution for x(x+3)+1=3(x+x^2) equation:

x(x+3)+1=3(x+x^2)

We move all terms to the left:

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

We multiply parentheses

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


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

3(x+x^2)

We multiply parentheses

3x^2+3x

Back to the equation:

-(3x^2+3x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-3x^2+3x-3x+1=0

We add all the numbers together, and all the variables

-2x^2+1=0

a = -2; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-2)·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-2}=\frac{0-2\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-4}
=-\frac{\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-2}=\frac{0+2\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-4}
=\frac{\sqrt{2}}{-2}
$