Exercise 25 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. 10x^2 = - 7x + 6 ⇕ 10x^2 + 7x - 6 = 0 Now, we can identify the values of a, b, and c. 10x^2 + 7x - 6=0 ⇕ 10x^2+ 7x+( - 6)=0 We see that a= 10, b= 7, and c= - 6. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 7±sqrt(7^2-4( 10)( - 6))/2( 10)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 7±sqrt(49-4(10)(- 6))/2(10)

Multiply

Multiply

x=- 7±sqrt(49-40(- 6))/20

MultNegNegOnePar

- a(- b)=a* b

x=- 7±sqrt(49+240)/20

AddTerms

Add terms

x=- 7±sqrt(289)/20

CalcRoot

Calculate root

x=- 7 ± 17/20

The solutions for this equation are x= - 7 ± 1720. Let's separate them into the positive and negative cases.

x=- 7± 17/20
x_1=- 7 + 17/20	x_2=- 7 - 17/20
x_1=10/20	x_2=- 24/20
x_1 = 0.5	x_2 = - 1.2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=0.5 and x_2=-1.2.