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10-3x+x^2-6x=90
We move all terms to the left:
10-3x+x^2-6x-(90)=0
We add all the numbers together, and all the variables
x^2-9x-80=0
a = 1; b = -9; c = -80;
Δ = b2-4ac
Δ = -92-4·1·(-80)
Δ = 401
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{401}}{2*1}=\frac{9-\sqrt{401}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{401}}{2*1}=\frac{9+\sqrt{401}}{2} $
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