Exercise 55 Page 767

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 and then simplify as much as possible.

28 = x^2+3x

SubEqn

LHS-28=RHS-28

0 = x^2+3x - 28

RearrangeEqn

Rearrange equation

x^2+3x - 28 = 0

Now, we can identify the values of a, b, and c. x^2+3x-28=0 ⇕ 1x^2+ 3x+( - 28)=0 We see that a= 1, b= 3, and c= - 28. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 3±sqrt(3^2-4( 1)( - 28))/2( 1)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 3±sqrt(9-4(1)(- 28))/2(1)

Multiply

Multiply

x=- 3±sqrt(9+112)/2

AddTerms

Add terms

x=- 3±sqrt(121)/2

CalcRoot

Calculate root

x = - 3 ± 11/2

The solutions for this equation are x= - 3 ± 112. Let's separate them into the positive and negative cases.

x=- 3± 11/2
x_1=- 3 + 11/2	x_2=- 3 - 11/2
x_1=8/2	x_2=- 14/2
x_1= 4	x_2=- 7

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=4 and x_2=- 7.