Answer:
The solution is [tex]\displaystyle x=1\pm \sqrt{47}[/tex]. Fourth option
Explanation:
Solve for x:
[tex]2x^2+3x-7=x^2+5x+39[/tex]
Move all the terms from the right to the left side of the equation, a zero in the right side:
[tex]2x^2+3x-7-x^2-5x-39=0[/tex]
Join all like terms:
[tex]x^2-2x-46=0[/tex]
The general form of the quadratic equation is:
[tex]ax^2+bx+c=0[/tex]
Solve the quadratic equation by using the formula:
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
In our equation: a=1, b=-2, c=-46
Substituting into the formula:
[tex]\displaystyle x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-46)}}{2(1)}[/tex]
[tex]\displaystyle x=\frac{2\pm \sqrt{4+184}}{2}[/tex]
[tex]\displaystyle x=\frac{2\pm \sqrt{188}}{2}[/tex]
Since 188=4*47
[tex]\displaystyle x=\frac{2\pm \sqrt{4*47}}{2}[/tex]
Take the square root of 4:
[tex]\displaystyle x=\frac{2\pm 2\sqrt{47}}{2}[/tex]
Divide by 2:
[tex]\displaystyle x=1\pm \sqrt{47}[/tex]
First option: Incorrect. The answer does not match
Second option: Incorrect. The answer does not match
Third option: Incorrect. The answer does not match
Fourth option: Correct. The answer matches exactly this option
