What are the solutions to the equation w^2-13w+42=0 ?

The solutions to the equation w^2-13w+42=0 are w=7,w=6

Find the zeros of w^2-13w+42=0

The zeros of w^2-13w+42=0 are w=7,w=6

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

w = 7, w = 6

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

w = 7, w = 6

\frac{960 ( 7 )}{( 7 ^{2} + 4 ) ^{2}}

6 x^{2} - 6 x + 1 = 0

\frac{m ^{2}}{324} = \frac{1}{2}

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

\frac{m - 2}{3} < - 2

What are the solutions to the equation w^2-13w+42=0 ?
The solutions to the equation w^2-13w+42=0 are w=7,w=6
Find the zeros of w^2-13w+42=0
The zeros of w^2-13w+42=0 are w=7,w=6