What are the solutions to the equation x^2+3x-7+8x=4 ?

The solutions to the equation x^2+3x-7+8x=4 are x=(-11+sqrt(165))/2 ,x=(-11-sqrt(165))/2

Find the zeros of x^2+3x-7+8x=4

The zeros of x^2+3x-7+8x=4 are x=(-11+sqrt(165))/2 ,x=(-11-sqrt(165))/2

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x^{2} + 3 x - 7 + 8 x = 4

x = \frac{- 11 + 165}{2}, x = \frac{- 11 - 165}{2}

Decimal Notation

x = 0.92261 \dots, x = - 11.92261 \dots

Solution steps

x^{2} + 3 x - 7 + 8 x = 4

Refine

x^{2} + 11 x - 7 = 4

Move

4

to the left side

x^{2} + 11 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot (- 11) = 165

x_{1, 2} = \frac{- 11 \pm 165}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 11 + 165}{2 \cdot 1}, x_{2} = \frac{- 11 - 165}{2 \cdot 1}

x = \frac{- 11 + 165}{2 \cdot 1} : \frac{- 11 + 165}{2}

x = \frac{- 11 - 165}{2 \cdot 1} : \frac{- 11 - 165}{2}

The solutions to the quadratic equation are:

x = \frac{- 11 + 165}{2}, x = \frac{- 11 - 165}{2}

\frac{1}{2} \cdot 9 \cdot 9

\frac{x}{4} = - 16

9 y^{2} - 3 y

8 > ∣ 6 x - 1 ∣ > 7

(\frac{x ^{- 11} y ^{- 2}}{z ^{8}})^{- 4}

What are the solutions to the equation x^2+3x-7+8x=4 ?
The solutions to the equation x^2+3x-7+8x=4 are x=(-11+sqrt(165))/2 ,x=(-11-sqrt(165))/2
Find the zeros of x^2+3x-7+8x=4
The zeros of x^2+3x-7+8x=4 are x=(-11+sqrt(165))/2 ,x=(-11-sqrt(165))/2