a^2+(3a^2)=1000

Solution for a^2+(3a^2)=1000 equation:

a^2+(3a^2)=1000

We move all terms to the left:

a^2+(3a^2)-(1000)=0

We add all the numbers together, and all the variables

4a^2-1000=0

a = 4; b = 0; c = -1000;
Δ = b2-4ac
Δ = 02-4·4·(-1000)
Δ = 16000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16000}=\sqrt{1600*10}=\sqrt{1600}*\sqrt{10}=40\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{10}}{2*4}=\frac{0-40\sqrt{10}}{8}
=-\frac{40\sqrt{10}}{8}
=-5\sqrt{10}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{10}}{2*4}=\frac{0+40\sqrt{10}}{8}
=\frac{40\sqrt{10}}{8}
=5\sqrt{10}
$