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2b^2-8=-4b
We move all terms to the left:
2b^2-8-(-4b)=0
We get rid of parentheses
2b^2+4b-8=0
a = 2; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·2·(-8)
Δ = 80
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{5}}{2*2}=\frac{-4-4\sqrt{5}}{4} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{5}}{2*2}=\frac{-4+4\sqrt{5}}{4} $
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