What are the solutions to the equation (1-3(-1))x^2+3x-4=0 ?

The solutions to the equation (1-3(-1))x^2+3x-4=0 are x=(-3+sqrt(73))/8 ,x=(-3-sqrt(73))/8

Find the zeros of (1-3(-1))x^2+3x-4=0

The zeros of (1-3(-1))x^2+3x-4=0 are x=(-3+sqrt(73))/8 ,x=(-3-sqrt(73))/8

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(1 - 3 (- 1)) x^{2} + 3 x - 4 = 0

x = \frac{- 3 + 73}{8}, x = \frac{- 3 - 73}{8}

Decimal Notation

x = 0.69300 \dots, x = - 1.44300 \dots

Solution steps

(1 - 3 (- 1)) x^{2} + 3 x - 4 = 0

Expand

(1 - 3 (- 1)) x^{2} + 3 x - 4 : 4 x^{2} + 3 x - 4

4 x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 4 (- 4) = 73

x_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 73}{2 \cdot 4}, x_{2} = \frac{- 3 - 73}{2 \cdot 4}

x = \frac{- 3 + 73}{2 \cdot 4} : \frac{- 3 + 73}{8}

x = \frac{- 3 - 73}{2 \cdot 4} : \frac{- 3 - 73}{8}

The solutions to the quadratic equation are:

x = \frac{- 3 + 73}{8}, x = \frac{- 3 - 73}{8}

x, 6 x^{2} - 9 x + 1 = 0

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

81 x^{2} - 5 = 0

1 0^{2} = 6^{2} + b^{2}

- 4 t^{2} - 28 t + 3 = 0

What are the solutions to the equation (1-3(-1))x^2+3x-4=0 ?
The solutions to the equation (1-3(-1))x^2+3x-4=0 are x=(-3+sqrt(73))/8 ,x=(-3-sqrt(73))/8
Find the zeros of (1-3(-1))x^2+3x-4=0
The zeros of (1-3(-1))x^2+3x-4=0 are x=(-3+sqrt(73))/8 ,x=(-3-sqrt(73))/8