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d^2=62
We move all terms to the left:
d^2-(62)=0
a = 1; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·1·(-62)
Δ = 248
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*1}=\frac{0-2\sqrt{62}}{2} =-\frac{2\sqrt{62}}{2} =-\sqrt{62} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*1}=\frac{0+2\sqrt{62}}{2} =\frac{2\sqrt{62}}{2} =\sqrt{62} $
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