a+a(a-3)=(a+4)-(a+1)

Solution for a+a(a-3)=(a+4)-(a+1) equation:

a+a(a-3)=(a+4)-(a+1)

We move all terms to the left:

a+a(a-3)-((a+4)-(a+1))=0

We multiply parentheses

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


We calculate terms in parentheses: -((a+4)-(a+1)), so:

(a+4)-(a+1)

We get rid of parentheses

a-a+4-1

We add all the numbers together, and all the variables

3

Back to the equation:

-(3)


We add all the numbers together, and all the variables

a^2-2a-3=0

a = 1; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*1}=\frac{-2}{2}
=-1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2}
=3
$