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4y^2+16y-560=0
a = 4; b = 16; c = -560;
Δ = b2-4ac
Δ = 162-4·4·(-560)
Δ = 9216
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{9216}=96$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-96}{2*4}=\frac{-112}{8} =-14 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+96}{2*4}=\frac{80}{8} =10 $
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