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

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

x - 5 \leq 4 x + 7

9 = \frac{m}{2}

\frac{2}{5} x + \frac{6}{5} = - \frac{2}{5}

x^{3} - 4 x^{2} > - 4 x + 16

- 5 x^{2} - 70 x - 65

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