k^2-14k+25=0=

Solution for k^2-14k+25=0= equation:

k^2-14k+25=0=

a = 1; b = -14; c = +25;
Δ = b2-4ac
Δ = -142-4·1·25
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{6}}{2*1}=\frac{14-4\sqrt{6}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{6}}{2*1}=\frac{14+4\sqrt{6}}{2}
$