Question

question 4 of 10
use the quadratic formula to find both solutions to the quadratic equation
given below.
$x^2 + 8x = 20$

a. $x = -10$
b. $x = \frac{3 - \sqrt{-7}}{2}$
c. $x = \frac{-6 + \sqrt{27}}{2}$
d. $x = -1$
e. $x = 2$
f. $x = \frac{-6 - \sqrt{27}}{2}$

Explanation:

Step1: Rewrite the equation in standard form

The standard form of a quadratic equation is \(ax^{2}+bx + c = 0\). Given \(x^{2}+8x=20\), we subtract 20 from both sides to get \(x^{2}+8x - 20=0\). Here, \(a = 1\), \(b = 8\), and \(c=- 20\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b = 8\), and \(c=-20\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(8)^{2}-4\times1\times(-20)=64 + 80 = 144\).
Then, \(x=\frac{-8\pm\sqrt{144}}{2\times1}=\frac{-8\pm12}{2}\).

Step3: Find the two solutions

For the plus sign: \(x=\frac{-8 + 12}{2}=\frac{4}{2}=2\).
For the minus sign: \(x=\frac{-8-12}{2}=\frac{-20}{2}=-10\).

Answer:

A. \(x = - 10\), E. \(x = 2\)