m2+20m+97=0

Solution for m2+20m+97=0 equation:

m2+20m+97=0

We add all the numbers together, and all the variables

m^2+20m+97=0

a = 1; b = 20; c = +97;
Δ = b2-4ac
Δ = 202-4·1·97
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{3}}{2*1}=\frac{-20-2\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{3}}{2*1}=\frac{-20+2\sqrt{3}}{2}
$