[tex]\begin{gathered} \text{ Given} \\ 4x^2+12x=-12 \end{gathered}[/tex]
First, equate the given to zero, and determine its coefficients a,b, and c
[tex]\begin{gathered} 4x^2+12x=-12 \\ 4x^2+12x+12=-12+12 \\ 4x^2+12x+12=0 \end{gathered}[/tex]
It is now in the standard form where a = 4, b = 12, and c = 12. Substitute these values to the quadratic formula and we get the following
[tex]\begin{gathered} x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \\ x = \frac{ -12 \pm \sqrt{12^2 - 4(4)(12)}}{ 2(4) } \\ x = \frac{ -12 \pm \sqrt{144 - 192}}{ 8 } \\ x = \frac{ -12 \pm \sqrt{-48}}{ 8 } \\ x = \frac{ -12 \pm 4\sqrt{3}\, i}{ 8 } \\ \; \\ x_1=\frac{ -12 }{ 8 }+\frac{4\sqrt{3}\, i}{ 8 } \\ x_1=-\frac{ 3}{ 2 }\pm\frac{ \sqrt{3}\, i}{ 2 } \\ \; \\ x_2=\frac{ -12 }{ 8 }-\frac{4\sqrt{3}\, i}{ 8 } \\ x_2=-\frac{ 3}{ 2 }-\frac{ \sqrt{3}\, i}{ 2 } \end{gathered}[/tex]
Therefore, the solution to the equation is
[tex]\begin{gathered} x=-\frac{ 3}{ 2 }+\frac{ \sqrt{3}\, i}{ 2 } \\ \text{ and} \\ x=-\frac{ 3}{ 2 }-\frac{ \sqrt{3}\, i}{ 2 } \end{gathered}[/tex]
