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3x^2+4x+12x=-2
We move all terms to the left:
3x^2+4x+12x-(-2)=0
We add all the numbers together, and all the variables
3x^2+16x+2=0
a = 3; b = 16; c = +2;
Δ = b2-4ac
Δ = 162-4·3·2
Δ = 232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{58}}{2*3}=\frac{-16-2\sqrt{58}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{58}}{2*3}=\frac{-16+2\sqrt{58}}{6} $
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