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

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

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

We multiply parentheses ..

(+8k^2+6k+12k+9)=0

We get rid of parentheses

8k^2+6k+12k+9=0

We add all the numbers together, and all the variables

8k^2+18k+9=0

a = 8; b = 18; c = +9;
Δ = b2-4ac
Δ = 182-4·8·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*8}=\frac{-24}{16}
=-1+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*8}=\frac{-12}{16}
=-3/4
$