k(-28k+15+1)=0

Solution for k(-28k+15+1)=0 equation:

k(-28k+15+1)=0

We add all the numbers together, and all the variables

k(-28k+16)=0

We multiply parentheses

-28k^2+16k=0

a = -28; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-28)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-28}=\frac{-32}{-56}
=4/7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-28}=\frac{0}{-56}
=0
$