3x^2+49=432

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Solution for 3x^2+49=432 equation:



3x^2+49=432
We move all terms to the left:
3x^2+49-(432)=0
We add all the numbers together, and all the variables
3x^2-383=0
a = 3; b = 0; c = -383;
Δ = b2-4ac
Δ = 02-4·3·(-383)
Δ = 4596
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{4596}=\sqrt{4*1149}=\sqrt{4}*\sqrt{1149}=2\sqrt{1149}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1149}}{2*3}=\frac{0-2\sqrt{1149}}{6} =-\frac{2\sqrt{1149}}{6} =-\frac{\sqrt{1149}}{3} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1149}}{2*3}=\frac{0+2\sqrt{1149}}{6} =\frac{2\sqrt{1149}}{6} =\frac{\sqrt{1149}}{3} $

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