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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

2(2x^2+1)

We multiply parentheses

4x^2+2

Back to the equation:

-(4x^2+2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+x+2=0

a = -3; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-3)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-3}=\frac{4}{-6}
=-2/3
$