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3x^2+18x-30=0
a = 3; b = 18; c = -30;
Δ = b2-4ac
Δ = 182-4·3·(-30)
Δ = 684
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{19}}{2*3}=\frac{-18-6\sqrt{19}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{19}}{2*3}=\frac{-18+6\sqrt{19}}{6} $
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