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x^2+56x+160=0
a = 1; b = 56; c = +160;
Δ = b2-4ac
Δ = 562-4·1·160
Δ = 2496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-8\sqrt{39}}{2*1}=\frac{-56-8\sqrt{39}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+8\sqrt{39}}{2*1}=\frac{-56+8\sqrt{39}}{2} $
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