What are the solutions to the equation 2w^2-12w+18=0 ?

The solutions to the equation 2w^2-12w+18=0 are w=3

Find the zeros of 2w^2-12w+18=0

The zeros of 2w^2-12w+18=0 are w=3

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2 w^{2} - 12 w + 18 = 0

w = 3

Solution steps

2 w^{2} - 12 w + 18 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 \cdot 18 = 0

w_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 2}

w = \frac{- ( - 12 )}{2 \cdot 2}

\frac{- ( - 12 )}{2 \cdot 2} = 3

w = 3

The solution to the quadratic equation is:

w = 3

x^{2} + 8 x + 15 = 0

y, x^{2} + y^{2} + 1 = 0

3 x^{2} + 2 x + 1

2 (x - 3)^{2} + 19 = 37

4 x^{2} + 12 x = - 1

What are the solutions to the equation 2w^2-12w+18=0 ?
The solutions to the equation 2w^2-12w+18=0 are w=3
Find the zeros of 2w^2-12w+18=0
The zeros of 2w^2-12w+18=0 are w=3