Respuesta :
First
[tex] 4x^{2} + 12x + 3 = 0 [/tex]
determine if this equation were factorable, it's not for this one
so I use quadratic formula, try to search the formula online
[tex]x1 = \frac{-12 + \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 + 4\sqrt{6} }{8} = \frac{4(-3 + \sqrt{6}) }{8} = \frac{-3 + \sqrt{6}}{2} = - 0.28[/tex]
[tex]x2 = \frac{-12 - \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 - 4\sqrt{6} }{8} = \frac{4(-3 - \sqrt{6}) }{8} = \frac{-3 - \sqrt{6}}{2} = - 2.72[/tex]
x ≈ −2.72 and x ≈ −0.28 is the solution
[tex] 4x^{2} + 12x + 3 = 0 [/tex]
determine if this equation were factorable, it's not for this one
so I use quadratic formula, try to search the formula online
[tex]x1 = \frac{-12 + \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 + 4\sqrt{6} }{8} = \frac{4(-3 + \sqrt{6}) }{8} = \frac{-3 + \sqrt{6}}{2} = - 0.28[/tex]
[tex]x2 = \frac{-12 - \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 - 4\sqrt{6} }{8} = \frac{4(-3 - \sqrt{6}) }{8} = \frac{-3 - \sqrt{6}}{2} = - 2.72[/tex]
x ≈ −2.72 and x ≈ −0.28 is the solution
Answer:
-2.72, -0.28 is correct
(just confirming)
Step-by-step explanation:
