Answer:
The answer is
[tex]x_1 = \frac{ - 3 + \sqrt{39} }{2} \: \:, \: \: \: \: x _2 = \frac{ - 3 - \sqrt{39} }{2} \\ [/tex]
Step-by-step explanation:
15 - 6x - 2x² = 0
Rewriting it we have
2x² + 6x - 15 = 0
Using the quadratic formula that's
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ [/tex]
From the question
a = 2 , b = 6 , c = - 15
So we have
[tex]x = \frac{ - 6 \pm \sqrt{ {6}^{2} - 4(2)( - 15) } }{2(2)} \\ = \frac{ - 6\pm \sqrt{36 + 120} }{4} \\ = \frac{ - 6\pm \sqrt{156} }{4} \: \: \: \: \: \: \: \: \: \: \\ = \frac{ - 6\pm2 \sqrt{39} }{4} \: \: \: \: \: \: \: \: \: \\ = - \frac{6}{4} \pm \frac{2 \sqrt{39} }{4} \: \: \: \: \: \: \: \: \\ = - \frac{3}{2} \pm \frac{ \sqrt{39} }{2} \: \: \: \: \: \: \: \: \: \: \\ = \frac{ - 3\pm \sqrt{39} }{2} \: \: \: \: \: \: \: \: \: \: \: [/tex]
Separate the solutions
We have the final answer as
[tex]x_1 = \frac{ - 3 + \sqrt{39} }{2} \: \: \: \: \: \: x _2 = \frac{ - 3 - \sqrt{39} }{2} \\ [/tex]
Hope this helps you
