m^2+6m-32=0

Solution for m^2+6m-32=0 equation:

m^2+6m-32=0

a = 1; b = 6; c = -32;
Δ = b2-4ac
Δ = 62-4·1·(-32)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{41}}{2*1}=\frac{-6-2\sqrt{41}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{41}}{2*1}=\frac{-6+2\sqrt{41}}{2}
$