What are the solutions to the equation x^{(2)}+19x-42=0 ?

The solutions to the equation x^{(2)}+19x-42=0 are x=2,x=-21

Find the zeros of x^{(2)}+19x-42=0

The zeros of x^{(2)}+19x-42=0 are x=2,x=-21

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x^{(2)} + 19 x - 42 = 0

x = 2, x = - 21

Solution steps

x^{(2)} + 19 x - 42 = 0

Refine

x^{2} + 19 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1 9^{2} - 4 \cdot 1 \cdot (- 42) = 23

x_{1, 2} = \frac{- 19 \pm 23}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 19 + 23}{2 \cdot 1}, x_{2} = \frac{- 19 - 23}{2 \cdot 1}

x = \frac{- 19 + 23}{2 \cdot 1} : 2

x = \frac{- 19 - 23}{2 \cdot 1} : - 21

The solutions to the quadratic equation are:

x = 2, x = - 21

w^{2} = 144

x^{2} + 1 5^{2} = 2 0^{2}

5 x^{2} + 3 x - 7 = 0

- 2 x (x + 1) = 2 (x - 1)

0 = 3 x^{2} - 4 x + 2

What are the solutions to the equation x^{(2)}+19x-42=0 ?
The solutions to the equation x^{(2)}+19x-42=0 are x=2,x=-21
Find the zeros of x^{(2)}+19x-42=0
The zeros of x^{(2)}+19x-42=0 are x=2,x=-21