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3y^2+y-1000=0
a = 3; b = 1; c = -1000;
Δ = b2-4ac
Δ = 12-4·3·(-1000)
Δ = 12001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{12001}}{2*3}=\frac{-1-\sqrt{12001}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{12001}}{2*3}=\frac{-1+\sqrt{12001}}{6} $
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