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w(w+2)=1225
We move all terms to the left:
w(w+2)-(1225)=0
We multiply parentheses
w^2+2w-1225=0
a = 1; b = 2; c = -1225;
Δ = b2-4ac
Δ = 22-4·1·(-1225)
Δ = 4904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4904}=\sqrt{4*1226}=\sqrt{4}*\sqrt{1226}=2\sqrt{1226}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1226}}{2*1}=\frac{-2-2\sqrt{1226}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1226}}{2*1}=\frac{-2+2\sqrt{1226}}{2} $
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