What are the solutions to the equation x^{(2)}-13x+42=0 ?

The solutions to the equation x^{(2)}-13x+42=0 are x=7,x=6

Find the zeros of x^{(2)}-13x+42=0

The zeros of x^{(2)}-13x+42=0 are x=7,x=6

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x^{(2)} - 13 x + 42 = 0

x = 7, x = 6

Solution steps

x^{(2)} - 13 x + 42 = 0

Refine

x^{2} - 13 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 42 = 1

x_{1, 2} = \frac{- ( - 13 ) \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 13 ) - 1}{2 \cdot 1}

x = \frac{- ( - 13 ) + 1}{2 \cdot 1} : 7

x = \frac{- ( - 13 ) - 1}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 7, x = 6

(4 - 22) x^{2} - 1 = 0

2 k^{2} - 3 k - 5 = 0

x^{2} + 2 0^{2} = (36 - 10)^{2}

x^{2} - 4 x = 3

x^{2} - 4 x = 9

What are the solutions to the equation x^{(2)}-13x+42=0 ?
The solutions to the equation x^{(2)}-13x+42=0 are x=7,x=6
Find the zeros of x^{(2)}-13x+42=0
The zeros of x^{(2)}-13x+42=0 are x=7,x=6