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x^2-16x+63=3
We move all terms to the left:
x^2-16x+63-(3)=0
We add all the numbers together, and all the variables
x^2-16x+60=0
a = 1; b = -16; c = +60;
Δ = b2-4ac
Δ = -162-4·1·60
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*1}=\frac{12}{2} =6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*1}=\frac{20}{2} =10 $
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