Respuesta :
Answer:
The correct substitution of a, b, and c in the quadratic formula is given by
[tex]$ x=\frac{-12\pm\sqrt{(12)^2-4(4)(9)}}{2(4)} $[/tex]
[tex]x = - \frac{ 3}{2} \: and \: x = -\frac{ 3}{2} \\\\[/tex]
The solutions of the given quadratic equation are real and equal.
Step-by-step explanation:
The given quadratic equation is
[tex]4x^2+12x+9 = 0[/tex]
The coefficients a, b and c are as follow:
[tex]a = 4 \\\\b = 12\\\\c = 9[/tex]
The quadratic formula is given by
[tex]$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]
The correct substitution of a, b, and c in the quadratic formula is given by
[tex]$ x=\frac{-12\pm\sqrt{(12)^2-4(4)(9)}}{2(4)} $[/tex]
Bonus:
The solution of this quadratic equation is given by
[tex]x=\frac{-12\pm\sqrt{(144 - 144)}}{8} \\\\x=\frac{-12\pm\sqrt{0}}{8} \\\\x=\frac{-12\pm 0}{8} \\\\x=\frac{-12 + 0}{8} \: and \: x=\frac{-12 - 0}{8}\\\\x= -\frac{ 3}{2} \: and \: x = -\frac{ 3}{2} \\\\[/tex]
Therefore, the solutions of the given quadratic equation are real and equal.
