h^2-3h-36=0

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Solution for h^2-3h-36=0 equation:



h^2-3h-36=0
a = 1; b = -3; c = -36;
Δ = b2-4ac
Δ = -32-4·1·(-36)
Δ = 153
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:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{17}}{2*1}=\frac{3-3\sqrt{17}}{2} $
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{17}}{2*1}=\frac{3+3\sqrt{17}}{2} $

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