r2-26-56=0

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Solution for r2-26-56=0 equation:



r2-26-56=0
We add all the numbers together, and all the variables
r^2-82=0
a = 1; b = 0; c = -82;
Δ = b2-4ac
Δ = 02-4·1·(-82)
Δ = 328
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{82}}{2*1}=\frac{0-2\sqrt{82}}{2} =-\frac{2\sqrt{82}}{2} =-\sqrt{82} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{82}}{2*1}=\frac{0+2\sqrt{82}}{2} =\frac{2\sqrt{82}}{2} =\sqrt{82} $

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