What are the solutions to the equation 2x^2-20x+32=0 ?

The solutions to the equation 2x^2-20x+32=0 are x=8,x=2

Find the zeros of 2x^2-20x+32=0

The zeros of 2x^2-20x+32=0 are x=8,x=2

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2 x^{2} - 20 x + 32 = 0

x = 8, x = 2

Solution steps

2 x^{2} - 20 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 32 = 12

x_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 2}

x = \frac{- ( - 20 ) + 12}{2 \cdot 2} : 8

x = \frac{- ( - 20 ) - 12}{2 \cdot 2} : 2

The solutions to the quadratic equation are:

x = 8, x = 2

3 x^{2} - 10 x - 4 = 0

(2 x + 3) (x - 4) = 0

2 m^{2} - 32 = 0

r^{2} + r + 12 = 0

15 x - (11 x + 5) = - x^{2}

What are the solutions to the equation 2x^2-20x+32=0 ?
The solutions to the equation 2x^2-20x+32=0 are x=8,x=2
Find the zeros of 2x^2-20x+32=0
The zeros of 2x^2-20x+32=0 are x=8,x=2