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2x^2+5x-33=0
a = 2; b = 5; c = -33;
Δ = b2-4ac
Δ = 52-4·2·(-33)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-17}{2*2}=\frac{-22}{4} =-5+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+17}{2*2}=\frac{12}{4} =3 $
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