(3k+5)(k+1)=0

Solution for (3k+5)(k+1)=0 equation:

(3k+5)(k+1)=0

We multiply parentheses ..

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

We get rid of parentheses

3k^2+3k+5k+5=0

We add all the numbers together, and all the variables

3k^2+8k+5=0

a = 3; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·3·5
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*3}=\frac{-10}{6}
=-1+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*3}=\frac{-6}{6}
=-1
$