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-d^2=-12
We move all terms to the left:
-d^2-(-12)=0
We add all the numbers together, and all the variables
-1d^2+12=0
a = -1; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-1)·12
Δ = 48
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-1}=\frac{0-4\sqrt{3}}{-2} =-\frac{4\sqrt{3}}{-2} =-\frac{2\sqrt{3}}{-1} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-1}=\frac{0+4\sqrt{3}}{-2} =\frac{4\sqrt{3}}{-2} =\frac{2\sqrt{3}}{-1} $
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