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3x^2=1000
We move all terms to the left:
3x^2-(1000)=0
a = 3; b = 0; c = -1000;
Δ = b2-4ac
Δ = 02-4·3·(-1000)
Δ = 12000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12000}=\sqrt{400*30}=\sqrt{400}*\sqrt{30}=20\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{30}}{2*3}=\frac{0-20\sqrt{30}}{6} =-\frac{20\sqrt{30}}{6} =-\frac{10\sqrt{30}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{30}}{2*3}=\frac{0+20\sqrt{30}}{6} =\frac{20\sqrt{30}}{6} =\frac{10\sqrt{30}}{3} $
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