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

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

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

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

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

x = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

x = 3

The solution to the quadratic equation is:

x = 3

x^{2} = 54

- 2 - 9 x \geq - 2 + 10 x

x^{2} - 6 x = - 18

6 x^{2} - 5 x - 6 \geq 0

ln (- e)

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