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x^2+26x+56=0
a = 1; b = 26; c = +56;
Δ = b2-4ac
Δ = 262-4·1·56
Δ = 452
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{113}}{2*1}=\frac{-26-2\sqrt{113}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{113}}{2*1}=\frac{-26+2\sqrt{113}}{2} $
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