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6x^2+126x+270=0
a = 6; b = 126; c = +270;
Δ = b2-4ac
Δ = 1262-4·6·270
Δ = 9396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9396}=\sqrt{324*29}=\sqrt{324}*\sqrt{29}=18\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(126)-18\sqrt{29}}{2*6}=\frac{-126-18\sqrt{29}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(126)+18\sqrt{29}}{2*6}=\frac{-126+18\sqrt{29}}{12} $
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