20-3x2+5=9

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Solution for 20-3x2+5=9 equation:



20-3x^2+5=9
We move all terms to the left:
20-3x^2+5-(9)=0
We add all the numbers together, and all the variables
-3x^2+16=0
a = -3; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-3)·16
Δ = 192
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:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-3}=\frac{0-8\sqrt{3}}{-6} =-\frac{8\sqrt{3}}{-6} =-\frac{4\sqrt{3}}{-3} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-3}=\frac{0+8\sqrt{3}}{-6} =\frac{8\sqrt{3}}{-6} =\frac{4\sqrt{3}}{-3} $

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