What are the solutions to the equation 15x^2=40(x+2) ?

The solutions to the equation 15x^2=40(x+2) are x=4,x=-4/3

Find the zeros of 15x^2=40(x+2)

The zeros of 15x^2=40(x+2) are x=4,x=-4/3

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15 x^{2} = 40 (x + 2)

x = 4, x = - \frac{4}{3}

Decimal Notation

x = 4, x = - 1.33333 \dots

Solution steps

15 x^{2} = 40 (x + 2)

Expand

40 (x + 2) : 40 x + 80

15 x^{2} = 40 x + 80

Move

80

to the left side

15 x^{2} - 80 = 40 x

Move

40 x

to the left side

15 x^{2} - 80 - 40 x = 0

Write in the standard form

a x^{2} + b x + c = 0

15 x^{2} - 40 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 15 ( - 80 )}{2 \cdot 15}

(- 40)^{2} - 4 \cdot 15 (- 80) = 80

x_{1, 2} = \frac{- ( - 40 ) \pm 80}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 80}{2 \cdot 15}, x_{2} = \frac{- ( - 40 ) - 80}{2 \cdot 15}

x = \frac{- ( - 40 ) + 80}{2 \cdot 15} : 4

x = \frac{- ( - 40 ) - 80}{2 \cdot 15} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = 4, x = - \frac{4}{3}

12 x = 13 - x^{2}

6 k^{2} + 12 k - 15 = - 10

x, x^{2} = k y^{3}

(x + 6)^{2} = 40

18 x^{2} - 720 x + 5400 = 0

What are the solutions to the equation 15x^2=40(x+2) ?
The solutions to the equation 15x^2=40(x+2) are x=4,x=-4/3
Find the zeros of 15x^2=40(x+2)
The zeros of 15x^2=40(x+2) are x=4,x=-4/3