Exercise 41 Page 145

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+18x=10

SubEqn

LHS-10=RHS-10

4x^2+18x-10=0

Now, we can identify the values of a, b, and c. 4x^2+18x-10=0 ⇕ 4x^2+ 18x+( - 10)=0 We see that a= 4, b= 18, and c= - 10. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 18±sqrt(18^2-4( 4)( - 10))/2( 4)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=-18±sqrt(324-4(4)(- 10))/2(4)

Multiply

Multiply

x=-18±sqrt(324-16(- 10))/8

MultNegNegOnePar

- a(- b)=a* b

x=-18±sqrt(324+160)/8

AddTerms

Add terms

x=-18±sqrt(484)/8

CalcRoot

Calculate root

x=- 18± 22/8

FactorOut

Factor out 2

x=2(-9± 11)/8

CancelCommonFac

Cancel out common factors

x=-9 ± 11/4

Using the Quadratic Formula, we found that the solutions of the given equation are x= -9± 114. Therefore, the solutions are x_1= 24=0.5 and x_2= -204=-5.