What are the solutions to the equation x^2+4x=21 ?

The solutions to the equation x^2+4x=21 are x=3,x=-7

Find the zeros of x^2+4x=21

The zeros of x^2+4x=21 are x=3,x=-7

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x^{2} + 4 x = 21

x = 3, x = - 7

Solution steps

x^{2} + 4 x = 21

Move

21

to the left side

x^{2} + 4 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 21) = 10

x_{1, 2} = \frac{- 4 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 10}{2 \cdot 1}, x_{2} = \frac{- 4 - 10}{2 \cdot 1}

x = \frac{- 4 + 10}{2 \cdot 1} : 3

x = \frac{- 4 - 10}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 3, x = - 7

(- i) (- i)

\frac{4 p}{5} + 3 = \frac{3}{7} + 5 p

4 ∣ 7 x ∣ + 5 = 29

4 x^{2} - 29 x - 63 = 0

95 = - 4 + 33 x

What are the solutions to the equation x^2+4x=21 ?
The solutions to the equation x^2+4x=21 are x=3,x=-7
Find the zeros of x^2+4x=21
The zeros of x^2+4x=21 are x=3,x=-7