What are the solutions to the equation 1x^2+3x-28=0 ?

The solutions to the equation 1x^2+3x-28=0 are x=4,x=-7

Find the zeros of 1x^2+3x-28=0

The zeros of 1x^2+3x-28=0 are x=4,x=-7

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1 x^{2} + 3 x - 28 = 0

x = 4, x = - 7

Solution steps

1 \cdot x^{2} + 3 x - 28 = 0

Refine

x^{2} + 3 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

x_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 11}{2 \cdot 1}, x_{2} = \frac{- 3 - 11}{2 \cdot 1}

x = \frac{- 3 + 11}{2 \cdot 1} : 4

x = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 4, x = - 7

y^{2} - x^{2} + 2 x - 1

(\frac{7}{8})^{1.8}

- 5.71 t - (12.92 - 6.71 t) = - 14.83

2 π \cdot \frac{1}{2}

ln (\frac{1}{1 - x})

What are the solutions to the equation 1x^2+3x-28=0 ?
The solutions to the equation 1x^2+3x-28=0 are x=4,x=-7
Find the zeros of 1x^2+3x-28=0
The zeros of 1x^2+3x-28=0 are x=4,x=-7