What are the solutions to the equation x^{(2)}+7x-8=0 ?

The solutions to the equation x^{(2)}+7x-8=0 are x=1,x=-8

Find the zeros of x^{(2)}+7x-8=0

The zeros of x^{(2)}+7x-8=0 are x=1,x=-8

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x^{(2)} + 7 x - 8 = 0

x = 1, x = - 8

Solution steps

x^{(2)} + 7 x - 8 = 0

Refine

x^{2} + 7 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 8) = 9

x_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 9}{2 \cdot 1}, x_{2} = \frac{- 7 - 9}{2 \cdot 1}

x = \frac{- 7 + 9}{2 \cdot 1} : 1

x = \frac{- 7 - 9}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 1, x = - 8

3 x^{2} - 60 x + 100 = 0

6 x^{2} + 23 x = - 7

y^{2} - y - 2 y = 0

x, y^{2} - x^{2} + 1 = 50

x, x^{2} + 2 x + 1 = m^{2}

What are the solutions to the equation x^{(2)}+7x-8=0 ?
The solutions to the equation x^{(2)}+7x-8=0 are x=1,x=-8
Find the zeros of x^{(2)}+7x-8=0
The zeros of x^{(2)}+7x-8=0 are x=1,x=-8