Exercise 31 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. x^2 - 3x = 10 ⇔ x^2 - 3x - 10 = 0 Now, we can identify the values of a, b, and c. x^2 - 3x - 10 = 0 ⇕ 1x^2+( - 3)x+( - 10)=0 We see that a= 1, b= - 3, and c= - 10. 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)( - 10))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=3±sqrt((- 3)^2-4(1)(- 10))/2(1)

CalcPow

Calculate power

x=3±sqrt(9-4(1)(- 10))/2(1)

MultByOne

a * 1=a

x=3±sqrt(9-4(- 10))/2

MultNegNegOnePar

- a(- b)=a* b

x=3±sqrt(9 + 40)/2

AddTerms

Add terms

x=3±sqrt(49)/2

CalcRoot

Calculate root

x=3± 7/2

The solutions for this equation are x= 3± 72. Let's separate them into the positive and negative cases.

x=3± 7/2
x_1=3 - 7/2	x_2=3 + 7/2
x_1=- 4/2	x_2=10/2
x_1=- 2	x_2=5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=- 2 and x_2=5.