a1=3,a2=5

Solution for a1=3,a2=5 equation:

a1=3.a^2=5

We move all terms to the left:

a1-(3.a^2)=0

We add all the numbers together, and all the variables

a-(3.a^2)=0

We get rid of parentheses

-3.a^2+a=0

We add all the numbers together, and all the variables

-3a^2+a=0

a = -3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-3)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-3}=\frac{-2}{-6}
=1/3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-3}=\frac{0}{-6}
=0
$