What are the solutions to the equation x^2-18x+32=0 ?

The solutions to the equation x^2-18x+32=0 are x=16,x=2

Find the zeros of x^2-18x+32=0

The zeros of x^2-18x+32=0 are x=16,x=2

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x^{2} - 18 x + 32 = 0

x = 16, x = 2

Solution steps

x^{2} - 18 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 32 = 14

x_{1, 2} = \frac{- ( - 18 ) \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 14}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 14}{2 \cdot 1}

x = \frac{- ( - 18 ) + 14}{2 \cdot 1} : 16

x = \frac{- ( - 18 ) - 14}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 16, x = 2

x^{2} + 10 x = 39

1 - [\frac{- 5}{100} e^{- 5} - 0.01]

\frac{\frac{π}{2}}{5}

- 2 (2 x - 1) = 3 x

- 4 y^{3} - 116 y^{2} - 400 y

What are the solutions to the equation x^2-18x+32=0 ?
The solutions to the equation x^2-18x+32=0 are x=16,x=2
Find the zeros of x^2-18x+32=0
The zeros of x^2-18x+32=0 are x=16,x=2