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h^2+2h=36
We move all terms to the left:
h^2+2h-(36)=0
a = 1; b = 2; c = -36;
Δ = b2-4ac
Δ = 22-4·1·(-36)
Δ = 148
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{37}}{2*1}=\frac{-2-2\sqrt{37}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{37}}{2*1}=\frac{-2+2\sqrt{37}}{2} $
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