What are the solutions to the equation w^2-11w+28=0 ?

The solutions to the equation w^2-11w+28=0 are w=7,w=4

Find the zeros of w^2-11w+28=0

The zeros of w^2-11w+28=0 are w=7,w=4

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w^{2} - 11 w + 28 = 0

w = 7, w = 4

Solution steps

w^{2} - 11 w + 28 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

w_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, w_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

w = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

w = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

w = 7, w = 4

- 9 n^{2} + 6 = - 30

3 x^{2} - 2 x - 17 = - 1

(z - 6)^{2} = 2 z

5 x^{2} + 13 x - 46 = 0

x^{2} + b x = 0

What are the solutions to the equation w^2-11w+28=0 ?
The solutions to the equation w^2-11w+28=0 are w=7,w=4
Find the zeros of w^2-11w+28=0
The zeros of w^2-11w+28=0 are w=7,w=4