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63+x^2=16x
We move all terms to the left:
63+x^2-(16x)=0
a = 1; b = -16; c = +63;
Δ = b2-4ac
Δ = -162-4·1·63
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*1}=\frac{14}{2} =7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*1}=\frac{18}{2} =9 $
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