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3x^2-30x+54=0
a = 3; b = -30; c = +54;
Δ = b2-4ac
Δ = -302-4·3·54
Δ = 252
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{7}}{2*3}=\frac{30-6\sqrt{7}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{7}}{2*3}=\frac{30+6\sqrt{7}}{6} $
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