16(m^2+m-2)=0

Solution for 16(m^2+m-2)=0 equation:

16(m^2+m-2)=0

We multiply parentheses

16m^2+16m-32=0

a = 16; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·16·(-32)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-48}{2*16}=\frac{-64}{32}
=-2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+48}{2*16}=\frac{32}{32}
=1
$