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

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

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

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

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32 = x^{2} + 4 x

x = 4, x = - 8

Solution steps

32 = x^{2} + 4 x

Switch sides

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} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 8

u^{2} - 3 u - 18 = 0

3 x^{2} - 688 x + 25300 = 0

r = 4 r^{2} - 1

x^{2} + 2 x + 100 = 0

y^{2} + 11 y + 28 = 0

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