k^2-14k+27=0

Solution for k^2-14k+27=0 equation:

k^2-14k+27=0

a = 1; b = -14; c = +27;
Δ = b2-4ac
Δ = -142-4·1·27
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{22}}{2*1}=\frac{14-2\sqrt{22}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{22}}{2*1}=\frac{14+2\sqrt{22}}{2}
$