Exercise 19 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 - 8x = 6 ⇔ 5x^2 - 8x - 6 = 0 Now, we can identify the values of a, b, and c. 5x^2 - 8x - 6 = 0 ⇕ 5x^2+( - 8)x+( - 6)=0 We see that a= 5, b= - 8, and c= - 6. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -8)±sqrt(( - 8)^2-4( 5)( - 6))/2( 5)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=8±sqrt((- 8)^2-4(5)(- 6))/2(5)

CalcPow

Calculate power

x=8±sqrt(64-4(5)(- 6))/2(5)

Multiply

Multiply

x=8±sqrt(64-20(- 6))/10

MultNegNegOnePar

- a(- b)=a* b

x=8±sqrt(64+120)/10

AddTerms

Add terms

x=8±sqrt(184)/10

SplitIntoFactors

Split into factors

x=8±sqrt(4* 46)/10

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=8± sqrt(4)* sqrt(46)/10

CalcRoot

Calculate root

x=8 ± 2 sqrt(46)/10

FactorOut

Factor out 2

x=2(4± sqrt(46))/10

CancelCommonFac

Cancel out common factors

x=4± sqrt(46)/5

The solutions for this equation are x= 4± sqrt(46)5. Let's separate them into the positive and negative cases.

x=4± sqrt(46)/5
x_1=4 + sqrt(46)/5	x_2=4 - sqrt(46)/5
x_1 ≈ 2.2	x_2 ≈ - 0.6

Using the Quadratic Formula, we found that the solutions of the given equation are x_1 ≈ 2.2 and x_2 ≈ - 0.6.