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14+x^2-16x=0
a = 1; b = -16; c = +14;
Δ = b2-4ac
Δ = -162-4·1·14
Δ = 200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-10\sqrt{2}}{2*1}=\frac{16-10\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+10\sqrt{2}}{2*1}=\frac{16+10\sqrt{2}}{2} $
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