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10^2+h^2=26^2
We move all terms to the left:
10^2+h^2-(26^2)=0
We add all the numbers together, and all the variables
h^2-576=0
a = 1; b = 0; c = -576;
Δ = b2-4ac
Δ = 02-4·1·(-576)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*1}=\frac{-48}{2} =-24 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*1}=\frac{48}{2} =24 $
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