6a^2+1-5=0

Solution for 6a^2+1-5=0 equation:

6a^2+1-5=0

We add all the numbers together, and all the variables

6a^2-4=0

a = 6; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·6·(-4)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*6}=\frac{0-4\sqrt{6}}{12}
=-\frac{4\sqrt{6}}{12}
=-\frac{\sqrt{6}}{3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*6}=\frac{0+4\sqrt{6}}{12}
=\frac{4\sqrt{6}}{12}
=\frac{\sqrt{6}}{3}
$