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3x(2x+2)=54
We move all terms to the left:
3x(2x+2)-(54)=0
We multiply parentheses
6x^2+6x-54=0
a = 6; b = 6; c = -54;
Δ = b2-4ac
Δ = 62-4·6·(-54)
Δ = 1332
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{37}}{2*6}=\frac{-6-6\sqrt{37}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{37}}{2*6}=\frac{-6+6\sqrt{37}}{12} $
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