What are the solutions to the equation 4=x(x+3) ?

The solutions to the equation 4=x(x+3) are x=1,x=-4

Find the zeros of 4=x(x+3)

The zeros of 4=x(x+3) are x=1,x=-4

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4 = x (x + 3)

x = 1, x = - 4

Solution steps

4 = x (x + 3)

Expand

x (x + 3) : x^{2} + 3 x

4 = x^{2} + 3 x

Switch sides

x^{2} + 3 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

x = \frac{- 3 + 5}{2 \cdot 1} : 1

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

a = \frac{1}{2} π r^{2}

(5 x + n)^{2} - 4 = 0

e (7) = - x^{2} - 9 x

x^{2} = 1 7^{2} + 8^{2}

(3 x - 1) (2 x + 3) = x + 2 (x - 1)

What are the solutions to the equation 4=x(x+3) ?
The solutions to the equation 4=x(x+3) are x=1,x=-4
Find the zeros of 4=x(x+3)
The zeros of 4=x(x+3) are x=1,x=-4