Answer:
x = -3
Step-by-step explanation:
Given equation:
[tex]5x^2+bx+12=0[/tex]
This equation has known solution [tex]-\frac{4}{5},[/tex] this means that
[tex]5\cdot \left(-\dfrac{4}{5}\right)^2+b\cdot \left(-\dfrac{4}{5}\right)+12=0[/tex]
Solve it for b:
[tex]5\cdot \dfrac{16}{25}-\dfrac{4b}{5}+12=0\\ \\\dfrac{16}{5}-\dfrac{4b}{5}+12=0\\ \\16-4b+12\cdot 5=0\ \ [\text{Multiplied by 5}]\\ \\16-4b+60=0\\ \\-4b=-12-60\\ \\-4b=-76\\ \\4b=76\\ \\b=19[/tex]
Therefore, the equation is
[tex]5x^2+19x+12=0[/tex]
Solve it using quadratic formula:
[tex]D=b^2-4ac=19^2-4\cdot 5\cdot 12=361-240=121\\ \\\sqrt{d}=\sqrt{121}=11\\ \\x_{1,2}=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-19\pm 11}{2\cdot 5}=-3,\ -\dfrac{4}{5}[/tex]
The other solution is x = -3
