12r^2-7=0

Solution for 12r^2-7=0 equation:

12r^2-7=0

a = 12; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·12·(-7)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*12}=\frac{0-4\sqrt{21}}{24}
=-\frac{4\sqrt{21}}{24}
=-\frac{\sqrt{21}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*12}=\frac{0+4\sqrt{21}}{24}
=\frac{4\sqrt{21}}{24}
=\frac{\sqrt{21}}{6}
$