What are the solutions to the equation w^2-18w+56=0 ?

The solutions to the equation w^2-18w+56=0 are w=14,w=4

Find the zeros of w^2-18w+56=0

The zeros of w^2-18w+56=0 are w=14,w=4

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w^{2} - 18 w + 56 = 0

w = 14, w = 4

Solution steps

w^{2} - 18 w + 56 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 56 = 10

w_{1, 2} = \frac{- ( - 18 ) \pm 10}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 18 ) + 10}{2 \cdot 1}, w_{2} = \frac{- ( - 18 ) - 10}{2 \cdot 1}

w = \frac{- ( - 18 ) + 10}{2 \cdot 1} : 14

w = \frac{- ( - 18 ) - 10}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

w = 14, w = 4

x, 25 x^{2} + y^{2} - 100 x + 6 y + 190 = 0

6 v^{2} + 24 v = 0

(3 x + 2)^{2} = 12

- x^{2} + x = - 6

550 = 450 (1 + x)^{2}

What are the solutions to the equation w^2-18w+56=0 ?
The solutions to the equation w^2-18w+56=0 are w=14,w=4
Find the zeros of w^2-18w+56=0
The zeros of w^2-18w+56=0 are w=14,w=4