What are the solutions to the equation 5=x^2+4x ?

The solutions to the equation 5=x^2+4x are x=1,x=-5

Find the zeros of 5=x^2+4x

The zeros of 5=x^2+4x are x=1,x=-5

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5 = x^{2} + 4 x

x = 1, x = - 5

Solution steps

5 = x^{2} + 4 x

Switch sides

x^{2} + 4 x = 5

Move

5

to the left side

x^{2} + 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 6}{2 \cdot 1}, x_{2} = \frac{- 4 - 6}{2 \cdot 1}

x = \frac{- 4 + 6}{2 \cdot 1} : 1

x = \frac{- 4 - 6}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

6 x^{2} - 28 x + 24 = 0

- 0.01 x^{2} + 50 x + 300 = 0

(x - 2) (3 x + 1) = 0

n^{2} - 11 n + 24 = 0

a = 6 a^{2}

What are the solutions to the equation 5=x^2+4x ?
The solutions to the equation 5=x^2+4x are x=1,x=-5
Find the zeros of 5=x^2+4x
The zeros of 5=x^2+4x are x=1,x=-5