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33=2x^2-5x
We move all terms to the left:
33-(2x^2-5x)=0
We get rid of parentheses
-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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