m^2+4=55

Solution for m^2+4=55 equation:

m^2+4=55

We move all terms to the left:

m^2+4-(55)=0

We add all the numbers together, and all the variables

m^2-51=0

a = 1; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·1·(-51)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*1}=\frac{0-2\sqrt{51}}{2}
=-\frac{2\sqrt{51}}{2}
=-\sqrt{51}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*1}=\frac{0+2\sqrt{51}}{2}
=\frac{2\sqrt{51}}{2}
=\sqrt{51}
$