(2m+3)(4m+5)=0

Solution for (2m+3)(4m+5)=0 equation:

(2m+3)(4m+5)=0

We multiply parentheses ..

(+8m^2+10m+12m+15)=0

We get rid of parentheses

8m^2+10m+12m+15=0

We add all the numbers together, and all the variables

8m^2+22m+15=0

a = 8; b = 22; c = +15;
Δ = b2-4ac
Δ = 222-4·8·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*8}=\frac{-24}{16}
=-1+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*8}=\frac{-20}{16}
=-1+1/4
$