Answer:
[tex]\implies x=1\pm \sqrt{47}[/tex]
Step-by-step explanation:
We want to solve for x in [tex]2x^2+3x-7=x^2+5x+39[/tex]
You need to group and combine like terms and write in standard form:
[tex]2x^2-x^2+3x-5x-7-39=0[/tex]
[tex]\imples x^2-2x-46=0[/tex]
By comparing to [tex]ax^2+bx+c=0[/tex], we a=1,b=-2 and c=-46
The solution can be obtained using the quadratic formula.
[tex]x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
We substitute the coefficients to get:
[tex]x=\frac{--2\pm \sqrt{(-2)^2-4*1*-46} }{2*1}[/tex]
[tex]x=\frac{2\pm \sqrt{4+4*46} }{2}[/tex]
[tex]x=\frac{2\pm \sqrt{4*47} }{2}[/tex]
[tex]x=\frac{2\pm2\sqrt{47} }{2}[/tex]
[tex]\implies x=1\pm \sqrt{47}[/tex]
The last choice is correct
