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x^2+9^2=26^2
We move all terms to the left:
x^2+9^2-(26^2)=0
We add all the numbers together, and all the variables
x^2-595=0
a = 1; b = 0; c = -595;
Δ = b2-4ac
Δ = 02-4·1·(-595)
Δ = 2380
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2380}=\sqrt{4*595}=\sqrt{4}*\sqrt{595}=2\sqrt{595}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{595}}{2*1}=\frac{0-2\sqrt{595}}{2} =-\frac{2\sqrt{595}}{2} =-\sqrt{595} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{595}}{2*1}=\frac{0+2\sqrt{595}}{2} =\frac{2\sqrt{595}}{2} =\sqrt{595} $
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