(6k+5)(k-4)=0

Solution for (6k+5)(k-4)=0 equation:

(6k+5)(k-4)=0

We multiply parentheses ..

(+6k^2-24k+5k-20)=0

We get rid of parentheses

6k^2-24k+5k-20=0

We add all the numbers together, and all the variables

6k^2-19k-20=0

a = 6; b = -19; c = -20;
Δ = b2-4ac
Δ = -192-4·6·(-20)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*6}=\frac{-10}{12}
=-5/6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*6}=\frac{48}{12}
=4
$