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12x(3x+1)=7
We move all terms to the left:
12x(3x+1)-(7)=0
We multiply parentheses
36x^2+12x-7=0
a = 36; b = 12; c = -7;
Δ = b2-4ac
Δ = 122-4·36·(-7)
Δ = 1152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24\sqrt{2}}{2*36}=\frac{-12-24\sqrt{2}}{72} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24\sqrt{2}}{2*36}=\frac{-12+24\sqrt{2}}{72} $
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