(4k+3)(2k-5)=0

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

(4k+3)(2k-5)=0

We multiply parentheses ..

(+8k^2-20k+6k-15)=0

We get rid of parentheses

8k^2-20k+6k-15=0

We add all the numbers together, and all the variables

8k^2-14k-15=0

a = 8; b = -14; c = -15;
Δ = b2-4ac
Δ = -142-4·8·(-15)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*8}=\frac{-12}{16}
=-3/4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*8}=\frac{40}{16}
=2+1/2
$