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d^2=140
We move all terms to the left:
d^2-(140)=0
a = 1; b = 0; c = -140;
Δ = b2-4ac
Δ = 02-4·1·(-140)
Δ = 560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*1}=\frac{0-4\sqrt{35}}{2} =-\frac{4\sqrt{35}}{2} =-2\sqrt{35} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*1}=\frac{0+4\sqrt{35}}{2} =\frac{4\sqrt{35}}{2} =2\sqrt{35} $
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