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6x^2+3x+4=12
We move all terms to the left:
6x^2+3x+4-(12)=0
We add all the numbers together, and all the variables
6x^2+3x-8=0
a = 6; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·6·(-8)
Δ = 201
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{201}}{2*6}=\frac{-3-\sqrt{201}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{201}}{2*6}=\frac{-3+\sqrt{201}}{12} $
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