m^2+28m=1

Solution for m^2+28m=1 equation:

m^2+28m=1

We move all terms to the left:

m^2+28m-(1)=0

a = 1; b = 28; c = -1;
Δ = b2-4ac
Δ = 282-4·1·(-1)
Δ = 788
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:

$\sqrt{\Delta}=\sqrt{788}=\sqrt{4*197}=\sqrt{4}*\sqrt{197}=2\sqrt{197}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{197}}{2*1}=\frac{-28-2\sqrt{197}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{197}}{2*1}=\frac{-28+2\sqrt{197}}{2}
$