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4b^2+4b-1=0
a = 4; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·4·(-1)
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{-(4)-4\sqrt{2}}{2*4}=\frac{-4-4\sqrt{2}}{8} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2}}{2*4}=\frac{-4+4\sqrt{2}}{8} $
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