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y2+4y-4=0
We add all the numbers together, and all the variables
y^2+4y-4=0
a = 1; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·1·(-4)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2}}{2*1}=\frac{-4-4\sqrt{2}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2}}{2*1}=\frac{-4+4\sqrt{2}}{2} $
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