What are the solutions to the equation 2x^2-16x+32=0 ?

The solutions to the equation 2x^2-16x+32=0 are x=4

Find the zeros of 2x^2-16x+32=0

The zeros of 2x^2-16x+32=0 are x=4

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2 x^{2} - 16 x + 32 = 0

x = 4

Solution steps

2 x^{2} - 16 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

(- 16)^{2} - 4 \cdot 2 \cdot 32 = 0

x_{1, 2} = \frac{- ( - 16 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 16 )}{2 \cdot 2}

\frac{- ( - 16 )}{2 \cdot 2} = 4

x = 4

The solution to the quadratic equation is:

x = 4

120 + 0.75 x = 105 + 0.9 x

36 c^{2} - 25

\frac{3}{5} \cdot 15

31 \geq 20 + 5 (x - 2.8)

\frac{x}{- 5} < 4

What are the solutions to the equation 2x^2-16x+32=0 ?
The solutions to the equation 2x^2-16x+32=0 are x=4
Find the zeros of 2x^2-16x+32=0
The zeros of 2x^2-16x+32=0 are x=4