Exercise 6 Page 137

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Let's start by rewriting the equation so all of the terms are on the left-hand side. 5x^2 + 5 = - 13x ⇕ 5x^2 + 13x + 5 = 0 Now, we can identify the values of a, b, and c. 5x^2 + 13x + 5 = 0 ⇕ 5x^2+ 13x+ 5=0 We see that a= 5, b= 13, and c= 5. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 13±sqrt(( 13)^2-4( 5)( 5))/2( 5)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 13±sqrt(169-4(5)(5))/2(5)

Multiply

Multiply

x=- 13±sqrt(169-100)/10

SubTerm

Subtract term

x=- 13±sqrt(69)/10

The solutions for this equation are x= - 13 ± sqrt(69)10. Let's separate them into the positive and negative cases.

x=- 13 ± sqrt(69)/10
x_1=- 13 - sqrt(69)/10	x_2=- 13 + sqrt(69)/10
x_1 ≈ - 2.1	x_2 ≈ - 0.5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1 ≈ - 2.1 and x_2 ≈ - 0.5.