Answer :
Answer:
[tex]x = 3 - \sqrt{19} \: \: or \: \: x = 3 + \sqrt{19} [/tex]
Step-by-step explanation:
The given equation is
[tex] {x}^{2} - 2x - 10 = 4x[/tex]
We group the linear terms to get:
[tex] {x}^{2} - 2x - 4x - 10 = 0[/tex]
[tex] {x}^{2} - 6x - 10 = 0[/tex]
We compare to
[tex]a {x}^{2} + bx + c = 0[/tex]
This means
a=1, b=-6 and c=-10
We plug in the values into the quadratic formula:
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
This gives us
[tex]x = \frac{ - -6 \pm \sqrt{ {( - 6)}^{2} - 4 \times 1 \times - 10 } }{2 \times 1} [/tex]
[tex]x = \frac{ 6 \pm \sqrt{ 36 + 40 } }{2} [/tex]
[tex]x = \frac{ 6 \pm \sqrt{76 } }{2} [/tex]
[tex]x = \frac{ - 6 \pm 2\sqrt{19} }{2} [/tex]
[tex]x = 3 \pm \sqrt{19} [/tex]
The solutions are
[tex]x =3 + \sqrt{19} [/tex]
or
[tex]x = 3 - \sqrt{19} [/tex]