I5u^2-12=0

Solution for I5u^2-12=0 equation:

5I^2-12=0

a = 5; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·5·(-12)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*5}=\frac{0-4\sqrt{15}}{10}
=-\frac{4\sqrt{15}}{10}
=-\frac{2\sqrt{15}}{5}
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*5}=\frac{0+4\sqrt{15}}{10}
=\frac{4\sqrt{15}}{10}
=\frac{2\sqrt{15}}{5}
$