Question

solve for ( x ) in the equation ( 2x^2 + 3x - 7 = x^2 + 5x + 38 )
( \bigcirc ) ( x = -6 pm sqrt{82} )
( \bigcirc ) ( x = -6 pm 2sqrt{17} )
( \bigcirc ) ( x = 1 pm sqrt{33} )
( \bigcirc ) ( x = 1 pm sqrt{47} )

Explanation:

Step1: Rearrange to standard quadratic form

Subtract $x^2+5x+38$ from both sides:
$2x^2+3x-7 - x^2-5x-38 = 0$
Simplify to: $x^2 - 2x - 45 = 0$

Step2: Apply quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=1$, $b=-2$, $c=-45$:
$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-45)}}{2(1)}$

Step3: Calculate discriminant and simplify

Compute discriminant: $\sqrt{4 + 180}=\sqrt{184}=2\sqrt{47}$
Simplify expression: $x=\frac{2\pm2\sqrt{47}}{2}=1\pm\sqrt{47}$

Answer:

$x=1\pm\sqrt{47}$ (matches the fourth option)