What are the solutions to the equation 15x^2+32x-7=0 ?

The solutions to the equation 15x^2+32x-7=0 are x= 1/5 ,x=-7/3

Find the zeros of 15x^2+32x-7=0

The zeros of 15x^2+32x-7=0 are x= 1/5 ,x=-7/3

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15 x^{2} + 32 x - 7 = 0

x = \frac{1}{5}, x = - \frac{7}{3}

Decimal Notation

x = 0.2, x = - 2.33333 \dots

Solution steps

15 x^{2} + 32 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 15 ( - 7 )}{2 \cdot 15}

3 2^{2} - 4 \cdot 15 (- 7) = 38

x_{1, 2} = \frac{- 32 \pm 38}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- 32 + 38}{2 \cdot 15}, x_{2} = \frac{- 32 - 38}{2 \cdot 15}

x = \frac{- 32 + 38}{2 \cdot 15} : \frac{1}{5}

x = \frac{- 32 - 38}{2 \cdot 15} : - \frac{7}{3}

The solutions to the quadratic equation are:

x = \frac{1}{5}, x = - \frac{7}{3}

- x - 8 = - 4 x - 23

∣ 4 x + 5 ∣ = 7

24 - 6 y - 4 x + x y

7 x^{2} + 6 x - 1 = 0

- 9 (x - 5) + 9 < 3 x + 42

What are the solutions to the equation 15x^2+32x-7=0 ?
The solutions to the equation 15x^2+32x-7=0 are x= 1/5 ,x=-7/3
Find the zeros of 15x^2+32x-7=0
The zeros of 15x^2+32x-7=0 are x= 1/5 ,x=-7/3