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1200=3x^2
We move all terms to the left:
1200-(3x^2)=0
a = -3; b = 0; c = +1200;
Δ = b2-4ac
Δ = 02-4·(-3)·1200
Δ = 14400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{14400}=120$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-120}{2*-3}=\frac{-120}{-6} =+20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+120}{2*-3}=\frac{120}{-6} =-20 $
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