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102+x-3x^2=0
a = -3; b = 1; c = +102;
Δ = b2-4ac
Δ = 12-4·(-3)·102
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-35}{2*-3}=\frac{-36}{-6} =+6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+35}{2*-3}=\frac{34}{-6} =-5+2/3 $
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