Question

solve for x in the equation $2x^{2}+3x - 7=x^{2}+5x + 39$. $x=-6pmsqrt{82}$ $x=-6pm2sqrt{17}$ $x = 1pmsqrt{33}$ $x=1pmsqrt{47}$

Explanation:

Step1: Rearrange to standard quadratic form

Subtract $x^2 + 5x + 39$ from both sides:
$2x^2 + 3x - 7 - x^2 - 5x - 39 = 0$
Simplify to get: $x^2 - 2x - 46 = 0$

Step2: Identify quadratic coefficients

For $ax^2 + bx + c = 0$, we have:
$a=1$, $b=-2$, $c=-46$

Step3: Apply quadratic formula

Use $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-46)}}{2(1)}$

Step4: Compute discriminant and simplify

Calculate discriminant: $\sqrt{4 + 184} = \sqrt{188} = \sqrt{4 \times 47} = 2\sqrt{47}$
Simplify the formula:
$x = \frac{2 \pm 2\sqrt{47}}{2} = 1 \pm \sqrt{47}$

Answer:

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