3m2+13=m2+397

Solution for 3m2+13=m2+397 equation:

3m^2+13=m2+397

We move all terms to the left:

3m^2+13-(m2+397)=0

We add all the numbers together, and all the variables

3m^2-(+m^2+397)+13=0

We get rid of parentheses

3m^2-m^2-397+13=0

We add all the numbers together, and all the variables

2m^2-384=0

a = 2; b = 0; c = -384;
Δ = b2-4ac
Δ = 02-4·2·(-384)
Δ = 3072
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{3}}{2*2}=\frac{0-32\sqrt{3}}{4}
=-\frac{32\sqrt{3}}{4}
=-8\sqrt{3}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{3}}{2*2}=\frac{0+32\sqrt{3}}{4}
=\frac{32\sqrt{3}}{4}
=8\sqrt{3}
$