28k^2+67k=40

Solution for 28k^2+67k=40 equation:

28k^2+67k=40

We move all terms to the left:

28k^2+67k-(40)=0

a = 28; b = 67; c = -40;
Δ = b2-4ac
Δ = 672-4·28·(-40)
Δ = 8969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(67)-\sqrt{8969}}{2*28}=\frac{-67-\sqrt{8969}}{56}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+\sqrt{8969}}{2*28}=\frac{-67+\sqrt{8969}}{56}
$