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

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

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

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

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

x = 4

Solution steps

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

Move

32

to the left side

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

3 - y^{2} = y^{2} + 1

x^{2} + 0 x - 9 = 0

p^{2} + 3 p = 2

9 x^{2} + 18 x = 0

x^{2} + 16 = 81

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