k^2-2k-224=0

Solution for k^2-2k-224=0 equation:

k^2-2k-224=0

a = 1; b = -2; c = -224;
Δ = b2-4ac
Δ = -22-4·1·(-224)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-30}{2*1}=\frac{-28}{2}
=-14
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+30}{2*1}=\frac{32}{2}
=16
$