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4-4x+x^2=100
We move all terms to the left:
4-4x+x^2-(100)=0
We add all the numbers together, and all the variables
x^2-4x-96=0
a = 1; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·1·(-96)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*1}=\frac{-16}{2} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*1}=\frac{24}{2} =12 $
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