What are the solutions to the equation 40x^2+27x-4=0 ?

The solutions to the equation 40x^2+27x-4=0 are x= 1/8 ,x=-4/5

Find the zeros of 40x^2+27x-4=0

The zeros of 40x^2+27x-4=0 are x= 1/8 ,x=-4/5

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40 x^{2} + 27 x - 4 = 0

x = \frac{1}{8}, x = - \frac{4}{5}

Decimal Notation

x = 0.125, x = - 0.8

Solution steps

40 x^{2} + 27 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 \cdot 40 ( - 4 )}{2 \cdot 40}

2 7^{2} - 4 \cdot 40 (- 4) = 37

x_{1, 2} = \frac{- 27 \pm 37}{2 \cdot 40}

Separate the solutions

x_{1} = \frac{- 27 + 37}{2 \cdot 40}, x_{2} = \frac{- 27 - 37}{2 \cdot 40}

x = \frac{- 27 + 37}{2 \cdot 40} : \frac{1}{8}

x = \frac{- 27 - 37}{2 \cdot 40} : - \frac{4}{5}

The solutions to the quadratic equation are:

x = \frac{1}{8}, x = - \frac{4}{5}

x^{2} - 8 x + 19 = 0

12 x^{2} - 128 x + 240 = 0

5 x^{2} + 12 x + 4 = 0

2 x^{2} - 20 = 10

x^{2} + 4 x - 19 = 0

What are the solutions to the equation 40x^2+27x-4=0 ?
The solutions to the equation 40x^2+27x-4=0 are x= 1/8 ,x=-4/5
Find the zeros of 40x^2+27x-4=0
The zeros of 40x^2+27x-4=0 are x= 1/8 ,x=-4/5