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5-3x=2x^2
We move all terms to the left:
5-3x-(2x^2)=0
determiningTheFunctionDomain -2x^2-3x+5=0
a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4} =-2+1/2 $
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