(2m+3)(4m+)=-0

Solution for (2m+3)(4m+)=-0 equation:

(2m+3)(4m+)=-0

We add all the numbers together, and all the variables

(2m+3)(+4m)=0

We multiply parentheses ..

(+8m^2+12m)=0

We get rid of parentheses

8m^2+12m=0

a = 8; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·8·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*8}=\frac{-24}{16}
=-1+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*8}=\frac{0}{16}
=0
$