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600-12x^2=20x^2
We move all terms to the left:
600-12x^2-(20x^2)=0
We add all the numbers together, and all the variables
-32x^2+600=0
a = -32; b = 0; c = +600;
Δ = b2-4ac
Δ = 02-4·(-32)·600
Δ = 76800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76800}=\sqrt{25600*3}=\sqrt{25600}*\sqrt{3}=160\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{3}}{2*-32}=\frac{0-160\sqrt{3}}{-64} =-\frac{160\sqrt{3}}{-64} =-\frac{5\sqrt{3}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{3}}{2*-32}=\frac{0+160\sqrt{3}}{-64} =\frac{160\sqrt{3}}{-64} =\frac{5\sqrt{3}}{-2} $
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