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-3y+4=y^2
We move all terms to the left:
-3y+4-(y^2)=0
determiningTheFunctionDomain -y^2-3y+4=0
We add all the numbers together, and all the variables
-1y^2-3y+4=0
a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2} =-4 $
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