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8=b^2
We move all terms to the left:
8-(b^2)=0
We add all the numbers together, and all the variables
-1b^2+8=0
a = -1; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-1)·8
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-1}=\frac{0-4\sqrt{2}}{-2} =-\frac{4\sqrt{2}}{-2} =-\frac{2\sqrt{2}}{-1} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-1}=\frac{0+4\sqrt{2}}{-2} =\frac{4\sqrt{2}}{-2} =\frac{2\sqrt{2}}{-1} $
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