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y^2+28y+26=0
a = 1; b = 28; c = +26;
Δ = b2-4ac
Δ = 282-4·1·26
Δ = 680
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{680}=\sqrt{4*170}=\sqrt{4}*\sqrt{170}=2\sqrt{170}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{170}}{2*1}=\frac{-28-2\sqrt{170}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{170}}{2*1}=\frac{-28+2\sqrt{170}}{2} $
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