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x^2+100=32x
We move all terms to the left:
x^2+100-(32x)=0
a = 1; b = -32; c = +100;
Δ = b2-4ac
Δ = -322-4·1·100
Δ = 624
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{39}}{2*1}=\frac{32-4\sqrt{39}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{39}}{2*1}=\frac{32+4\sqrt{39}}{2} $
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