21a^2-5=436

Solution for 21a^2-5=436 equation:

21a^2-5=436

We move all terms to the left:

21a^2-5-(436)=0

We add all the numbers together, and all the variables

21a^2-441=0

a = 21; b = 0; c = -441;
Δ = b2-4ac
Δ = 02-4·21·(-441)
Δ = 37044
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:

$\sqrt{\Delta}=\sqrt{37044}=\sqrt{1764*21}=\sqrt{1764}*\sqrt{21}=42\sqrt{21}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{21}}{2*21}=\frac{0-42\sqrt{21}}{42}
=-\frac{42\sqrt{21}}{42}
=-\sqrt{21}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{21}}{2*21}=\frac{0+42\sqrt{21}}{42}
=\frac{42\sqrt{21}}{42}
=\sqrt{21}
$