What are the solutions to the equation x^2+10x-57=4 ?

The solutions to the equation x^2+10x-57=4 are x=-5+sqrt(86),x=-5-sqrt(86)

Find the zeros of x^2+10x-57=4

The zeros of x^2+10x-57=4 are x=-5+sqrt(86),x=-5-sqrt(86)

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x^{2} + 10 x - 57 = 4

x = - 5 + 86, x = - 5 - 86

Decimal Notation

x = 4.27361 \dots, x = - 14.27361 \dots

Solution steps

x^{2} + 10 x - 57 = 4

Move

4

to the left side

x^{2} + 10 x - 61 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 61 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 61) = 286

x_{1, 2} = \frac{- 10 \pm 2 86}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 86}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 86}{2 \cdot 1}

x = \frac{- 10 + 2 86}{2 \cdot 1} : - 5 + 86

x = \frac{- 10 - 2 86}{2 \cdot 1} : - 5 - 86

The solutions to the quadratic equation are:

x = - 5 + 86, x = - 5 - 86

x^{2} - 4 x = - 6

4 x^{2} = 16 x - 21

6 x^{2} - 13 x - 8 = 0

5 x^{2} - 80 x = 0

15 y^{2} = 20

What are the solutions to the equation x^2+10x-57=4 ?
The solutions to the equation x^2+10x-57=4 are x=-5+sqrt(86),x=-5-sqrt(86)
Find the zeros of x^2+10x-57=4
The zeros of x^2+10x-57=4 are x=-5+sqrt(86),x=-5-sqrt(86)