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2x^2+x-4=0
a = 2; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·2·(-4)
Δ = 33
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{-(1)-\sqrt{33}}{2*2}=\frac{-1-\sqrt{33}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*2}=\frac{-1+\sqrt{33}}{4} $
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