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3x^2+12x=-7
We move all terms to the left:
3x^2+12x-(-7)=0
We add all the numbers together, and all the variables
3x^2+12x+7=0
a = 3; b = 12; c = +7;
Δ = b2-4ac
Δ = 122-4·3·7
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{15}}{2*3}=\frac{-12-2\sqrt{15}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{15}}{2*3}=\frac{-12+2\sqrt{15}}{6} $
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