What are the solutions to the equation x^2-4x=32 ?

The solutions to the equation x^2-4x=32 are x=8,x=-4

Find the zeros of x^2-4x=32

The zeros of x^2-4x=32 are x=8,x=-4

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x^{2} - 4 x = 32

x = 8, x = - 4

Solution steps

x^{2} - 4 x = 32

Move

32

to the left side

x^{2} - 4 x - 32 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 32) = 12

x_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 1}

x = \frac{- ( - 4 ) + 12}{2 \cdot 1} : 8

x = \frac{- ( - 4 ) - 12}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 8, x = - 4

- x + 17 - 8 x = 9 - x

\frac{16 !}{( 16 - 4 ) !}

200 (0.5)^{\frac{5}{2}}

x^{2} + 4 x + 11 = 0

3 (x + 5) = 2 (3 x + 12)

What are the solutions to the equation x^2-4x=32 ?
The solutions to the equation x^2-4x=32 are x=8,x=-4
Find the zeros of x^2-4x=32
The zeros of x^2-4x=32 are x=8,x=-4