What are the solutions to the equation x^2+10x-51=8 ?

The solutions to the equation x^2+10x-51=8 are x=-5+2sqrt(21),x=-5-2sqrt(21)

Find the zeros of x^2+10x-51=8

The zeros of x^2+10x-51=8 are x=-5+2sqrt(21),x=-5-2sqrt(21)

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x^{2} + 10 x - 51 = 8

x = - 5 + 221, x = - 5 - 221

Solution steps

x^{2} + 10 x - 51 = 8

Move

8

to the left side

x^{2} + 10 x - 59 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 59 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 59) = 421

x_{1, 2} = \frac{- 10 \pm 4 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 4 21}{2 \cdot 1}, x_{2} = \frac{- 10 - 4 21}{2 \cdot 1}

x = \frac{- 10 + 4 21}{2 \cdot 1} : - 5 + 221

x = \frac{- 10 - 4 21}{2 \cdot 1} : - 5 - 221

The solutions to the quadratic equation are:

x = - 5 + 221, x = - 5 - 221

3 x^{2} - 6 x + 3

3 (2 x - 2) = 2 (3 x + 9)

x^{3} - 2 x^{2} + 7 x - 14

72 x^{2} - 12 x - 24

(3^{2})^{3}

What are the solutions to the equation x^2+10x-51=8 ?
The solutions to the equation x^2+10x-51=8 are x=-5+2sqrt(21),x=-5-2sqrt(21)
Find the zeros of x^2+10x-51=8
The zeros of x^2+10x-51=8 are x=-5+2sqrt(21),x=-5-2sqrt(21)