m^2=1m+56

Solution for m^2=1m+56 equation:

m^2=1m+56

We move all terms to the left:

m^2-(1m+56)=0

We add all the numbers together, and all the variables

m^2-(m+56)=0

We get rid of parentheses

m^2-m-56=0

We add all the numbers together, and all the variables

m^2-1m-56=0

a = 1; b = -1; c = -56;
Δ = b2-4ac
Δ = -12-4·1·(-56)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*1}=\frac{-14}{2}
=-7
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*1}=\frac{16}{2}
=8
$