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y^2+16y-192=0
a = 1; b = 16; c = -192;
Δ = b2-4ac
Δ = 162-4·1·(-192)
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*1}=\frac{-48}{2} =-24 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*1}=\frac{16}{2} =8 $
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