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

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

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

We multiply parentheses ..

(+2k^2+10k-3k-15)=0

We get rid of parentheses

2k^2+10k-3k-15=0

We add all the numbers together, and all the variables

2k^2+7k-15=0

a = 2; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·2·(-15)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*2}=\frac{-20}{4}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*2}=\frac{6}{4}
=1+1/2
$