3e2+12=21

Solution for 3e2+12=21 equation:

3e^2+12=21

We move all terms to the left:

3e^2+12-(21)=0

We add all the numbers together, and all the variables

3e^2-9=0

a = 3; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·3·(-9)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*3}=\frac{0-6\sqrt{3}}{6}
=-\frac{6\sqrt{3}}{6}
=-\sqrt{3}
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*3}=\frac{0+6\sqrt{3}}{6}
=\frac{6\sqrt{3}}{6}
=\sqrt{3}
$