k^2+4k-16=80

Solution for k^2+4k-16=80 equation:

k^2+4k-16=80

We move all terms to the left:

k^2+4k-16-(80)=0

We add all the numbers together, and all the variables

k^2+4k-96=0

a = 1; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*1}=\frac{-24}{2}
=-12
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*1}=\frac{16}{2}
=8
$