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x^2+36x-288=0
a = 1; b = 36; c = -288;
Δ = b2-4ac
Δ = 362-4·1·(-288)
Δ = 2448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12\sqrt{17}}{2*1}=\frac{-36-12\sqrt{17}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12\sqrt{17}}{2*1}=\frac{-36+12\sqrt{17}}{2} $
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