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2y^2+4y-16=0
a = 2; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·2·(-16)
Δ = 144
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*2}=\frac{-16}{4} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*2}=\frac{8}{4} =2 $
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