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36-x^2=3x+8
We move all terms to the left:
36-x^2-(3x+8)=0
We add all the numbers together, and all the variables
-1x^2-(3x+8)+36=0
We get rid of parentheses
-1x^2-3x-8+36=0
We add all the numbers together, and all the variables
-1x^2-3x+28=0
a = -1; b = -3; c = +28;
Δ = b2-4ac
Δ = -32-4·(-1)·28
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-1}=\frac{-8}{-2} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-1}=\frac{14}{-2} =-7 $
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