2k^2-22k+36=0

Solution for 2k^2-22k+36=0 equation:

2k^2-22k+36=0

a = 2; b = -22; c = +36;
Δ = b2-4ac
Δ = -222-4·2·36
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*2}=\frac{8}{4}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*2}=\frac{36}{4}
=9
$