a+3+a2=-1+3a+4

Solution for a+3+a2=-1+3a+4 equation:

a+3+a2=-1+3a+4

We move all terms to the left:

a+3+a2-(-1+3a+4)=0

We add all the numbers together, and all the variables

a+a2-(3a+3)+3=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

a^2+a-3a-3+3=0

We add all the numbers together, and all the variables

a^2-2a=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:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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