Exercise 15 Page 195

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.

4x^2-6=- 12x

AddEqn

LHS+12x=RHS+12x

4x^2+12x-6=0

FactorOut

Factor out 2

2(2x^2+6x-3)=0

DivEqn

.LHS /2.=.RHS /2.

2x^2+6x-3=0

Now, we can identify the values of a, b, and c. 2x^2+6x-3=0 ⇔ 2x^2+ 6x+( - 3)=0 We see that a= 2, b= 6, and c= - 3. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 6±sqrt(6^2-4( 2)( - 3))/2( 2)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 6±sqrt(36-4(2)(- 3))/2(2)

Multiply

Multiply

x=- 6±sqrt(36-8(- 3))/4

MultNegNegOnePar

- a(- b)=a* b

x=- 6±sqrt(36+24)/4

AddTerms

Add terms

x=- 6±sqrt(60)/4

SplitIntoFactors

Split into factors

x=- 6±sqrt(4* 15)/4

SqrtProd

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

x=- 6± sqrt(4)* sqrt(15)/4

CalcRoot

Calculate root

x=- 6± 2 sqrt(15)/4

FactorOut

Factor out 2

x=2(- 3± sqrt(15))/4

CancelCommonFac

Cancel out common factors

x=- 3± sqrt(15)/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= - 3± sqrt(15)2. Therefore, the solutions are x_1= - 3+ sqrt(15)2 and x_2= - 3- sqrt(15)2.