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40-3x=x^2
We move all terms to the left:
40-3x-(x^2)=0
determiningTheFunctionDomain -x^2-3x+40=0
We add all the numbers together, and all the variables
-1x^2-3x+40=0
a = -1; b = -3; c = +40;
Δ = b2-4ac
Δ = -32-4·(-1)·40
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*-1}=\frac{16}{-2} =-8 $
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