3v^2+12=93

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Solution for 3v^2+12=93 equation:



3v^2+12=93
We move all terms to the left:
3v^2+12-(93)=0
We add all the numbers together, and all the variables
3v^2-81=0
a = 3; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·3·(-81)
Δ = 972
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{3}}{2*3}=\frac{0-18\sqrt{3}}{6} =-\frac{18\sqrt{3}}{6} =-3\sqrt{3} $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{3}}{2*3}=\frac{0+18\sqrt{3}}{6} =\frac{18\sqrt{3}}{6} =3\sqrt{3} $

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