2k^2=225

Solution for 2k^2=225 equation:

2k^2=225

We move all terms to the left:

2k^2-(225)=0

a = 2; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·2·(-225)
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{2}}{2*2}=\frac{0-30\sqrt{2}}{4}
=-\frac{30\sqrt{2}}{4}
=-\frac{15\sqrt{2}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{2}}{2*2}=\frac{0+30\sqrt{2}}{4}
=\frac{30\sqrt{2}}{4}
=\frac{15\sqrt{2}}{2}
$