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3-3x-x^2=0
We add all the numbers together, and all the variables
-1x^2-3x+3=0
a = -1; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-1)·3
Δ = 21
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{21}}{2*-1}=\frac{3-\sqrt{21}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*-1}=\frac{3+\sqrt{21}}{-2} $
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