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2x^2+3x-2998=0
a = 2; b = 3; c = -2998;
Δ = b2-4ac
Δ = 32-4·2·(-2998)
Δ = 23993
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{23993}}{2*2}=\frac{-3-\sqrt{23993}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{23993}}{2*2}=\frac{-3+\sqrt{23993}}{4} $
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