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y^2+256y+1088=0
a = 1; b = 256; c = +1088;
Δ = b2-4ac
Δ = 2562-4·1·1088
Δ = 61184
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{61184}=\sqrt{256*239}=\sqrt{256}*\sqrt{239}=16\sqrt{239}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(256)-16\sqrt{239}}{2*1}=\frac{-256-16\sqrt{239}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(256)+16\sqrt{239}}{2*1}=\frac{-256+16\sqrt{239}}{2} $
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