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3-3(x^2)=0
a = -3; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-3)·3
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-3}=\frac{-6}{-6} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-3}=\frac{6}{-6} =-1 $
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