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2x^2-33x+45=0
a = 2; b = -33; c = +45;
Δ = b2-4ac
Δ = -332-4·2·45
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-27}{2*2}=\frac{6}{4} =1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+27}{2*2}=\frac{60}{4} =15 $
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