x^2+4=(x+1)+(x+3)

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

x^2+4=(x+1)+(x+3)

We move all terms to the left:

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


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

(x+1)+(x+3)

We get rid of parentheses

x+x+1+3

We add all the numbers together, and all the variables

2x+4

Back to the equation:

-(2x+4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2x=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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