What are the solutions to the equation 3x^2+8x-20=0 ?

The solutions to the equation 3x^2+8x-20=0 are x=(2(sqrt(19)-2))/3 ,x=-(2(2+sqrt(19)))/3

Find the zeros of 3x^2+8x-20=0

The zeros of 3x^2+8x-20=0 are x=(2(sqrt(19)-2))/3 ,x=-(2(2+sqrt(19)))/3

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3 x^{2} + 8 x - 20 = 0

x = \frac{2 ( 19 - 2 )}{3}, x = - \frac{2 ( 2 + 19 )}{3}

Decimal Notation

x = 1.57259 \dots, x = - 4.23926 \dots

Solution steps

3 x^{2} + 8 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

8^{2} - 4 \cdot 3 (- 20) = 419

x_{1, 2} = \frac{- 8 \pm 4 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 8 + 4 19}{2 \cdot 3}, x_{2} = \frac{- 8 - 4 19}{2 \cdot 3}

x = \frac{- 8 + 4 19}{2 \cdot 3} : \frac{2 ( 19 - 2 )}{3}

x = \frac{- 8 - 4 19}{2 \cdot 3} : - \frac{2 ( 2 + 19 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 19 - 2 )}{3}, x = - \frac{2 ( 2 + 19 )}{3}

m^{2} + 14 m + 49 = 0

y, x = y^{2} - 4

y, x = y^{2} - 2

x, 5 x^{2} + 7 x - 2 = 0

- 4.9 t^{2} + 2 = 0

What are the solutions to the equation 3x^2+8x-20=0 ?
The solutions to the equation 3x^2+8x-20=0 are x=(2(sqrt(19)-2))/3 ,x=-(2(2+sqrt(19)))/3
Find the zeros of 3x^2+8x-20=0
The zeros of 3x^2+8x-20=0 are x=(2(sqrt(19)-2))/3 ,x=-(2(2+sqrt(19)))/3