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100=3x^2
We move all terms to the left:
100-(3x^2)=0
a = -3; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-3)·100
Δ = 1200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*-3}=\frac{0-20\sqrt{3}}{-6} =-\frac{20\sqrt{3}}{-6} =-\frac{10\sqrt{3}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*-3}=\frac{0+20\sqrt{3}}{-6} =\frac{20\sqrt{3}}{-6} =\frac{10\sqrt{3}}{-3} $
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