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x^2+27x-135=0
a = 1; b = 27; c = -135;
Δ = b2-4ac
Δ = 272-4·1·(-135)
Δ = 1269
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1269}=\sqrt{9*141}=\sqrt{9}*\sqrt{141}=3\sqrt{141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{141}}{2*1}=\frac{-27-3\sqrt{141}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{141}}{2*1}=\frac{-27+3\sqrt{141}}{2} $
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