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x^2=440
We move all terms to the left:
x^2-(440)=0
a = 1; b = 0; c = -440;
Δ = b2-4ac
Δ = 02-4·1·(-440)
Δ = 1760
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1760}=\sqrt{16*110}=\sqrt{16}*\sqrt{110}=4\sqrt{110}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{110}}{2*1}=\frac{0-4\sqrt{110}}{2} =-\frac{4\sqrt{110}}{2} =-2\sqrt{110} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{110}}{2*1}=\frac{0+4\sqrt{110}}{2} =\frac{4\sqrt{110}}{2} =2\sqrt{110} $
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