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-3x^2-3x+200=0
a = -3; b = -3; c = +200;
Δ = b2-4ac
Δ = -32-4·(-3)·200
Δ = 2409
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{2409}}{2*-3}=\frac{3-\sqrt{2409}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{2409}}{2*-3}=\frac{3+\sqrt{2409}}{-6} $
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