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3x^2+6x-135=0
a = 3; b = 6; c = -135;
Δ = b2-4ac
Δ = 62-4·3·(-135)
Δ = 1656
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1656}=\sqrt{36*46}=\sqrt{36}*\sqrt{46}=6\sqrt{46}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{46}}{2*3}=\frac{-6-6\sqrt{46}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{46}}{2*3}=\frac{-6+6\sqrt{46}}{6} $
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