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

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

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

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

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

x = 16, x = 2

Solution steps

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

Switch sides

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

- 0.5 x^{2} + 1.05 x + 0.4 = - 8.5

6 = x^{2} + 5 x

27 t^{2} - 36 t + 12 = 0

\frac{x + 7}{6} = \frac{x ^{2} - 1}{3}

36 x^{2} + 39 x = 12

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