What are the solutions to the equation 15x^2=2(4-7x) ?

The solutions to the equation 15x^2=2(4-7x) are x= 2/5 ,x=-4/3

Find the zeros of 15x^2=2(4-7x)

The zeros of 15x^2=2(4-7x) are x= 2/5 ,x=-4/3

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15 x^{2} = 2 (4 - 7 x)

x = \frac{2}{5}, x = - \frac{4}{3}

Decimal Notation

x = 0.4, x = - 1.33333 \dots

Solution steps

15 x^{2} = 2 (4 - 7 x)

Expand

2 (4 - 7 x) : 8 - 14 x

15 x^{2} = 8 - 14 x

Move

14 x

to the left side

15 x^{2} + 14 x = 8

Move

8

to the left side

15 x^{2} + 14 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 15 ( - 8 )}{2 \cdot 15}

1 4^{2} - 4 \cdot 15 (- 8) = 26

x_{1, 2} = \frac{- 14 \pm 26}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- 14 + 26}{2 \cdot 15}, x_{2} = \frac{- 14 - 26}{2 \cdot 15}

x = \frac{- 14 + 26}{2 \cdot 15} : \frac{2}{5}

x = \frac{- 14 - 26}{2 \cdot 15} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = \frac{2}{5}, x = - \frac{4}{3}

9 x^{2} - 81 x = 0

n^{2} + 5 n + 6 = 0

m^{2} + 7 m + 10 = 0

4 x = x^{2} + 4

x^{2} - 3 x + 14 = 0

What are the solutions to the equation 15x^2=2(4-7x) ?
The solutions to the equation 15x^2=2(4-7x) are x= 2/5 ,x=-4/3
Find the zeros of 15x^2=2(4-7x)
The zeros of 15x^2=2(4-7x) are x= 2/5 ,x=-4/3