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y^2-4y-32=0.
a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*1}=\frac{-8}{2} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2} =8 $
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