3m^2+7=439

Solution for 3m^2+7=439 equation:

3m^2+7=439

We move all terms to the left:

3m^2+7-(439)=0

We add all the numbers together, and all the variables

3m^2-432=0

a = 3; b = 0; c = -432;
Δ = b2-4ac
Δ = 02-4·3·(-432)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*3}=\frac{-72}{6}
=-12
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*3}=\frac{72}{6}
=12
$