What are the solutions to the equation 4=x^2+10x ?

The solutions to the equation 4=x^2+10x are x=-5+sqrt(29),x=-5-sqrt(29)

Find the zeros of 4=x^2+10x

The zeros of 4=x^2+10x are x=-5+sqrt(29),x=-5-sqrt(29)

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4 = x^{2} + 10 x

x = - 5 + 29, x = - 5 - 29

Solution steps

4 = x^{2} + 10 x

Switch sides

x^{2} + 10 x = 4

Move

4

to the left side

x^{2} + 10 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 4) = 229

x_{1, 2} = \frac{- 10 \pm 2 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 29}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 29}{2 \cdot 1}

x = \frac{- 10 + 2 29}{2 \cdot 1} : - 5 + 29

x = \frac{- 10 - 2 29}{2 \cdot 1} : - 5 - 29

The solutions to the quadratic equation are:

x = - 5 + 29, x = - 5 - 29

2 x - 5 (x + 3) = x (x - 4)

3 x + 4 = 2 x^{2} - 7

407 x^{2} + 480 x - 69 = 0

(2 x + 5) (6 x - 1) = 120

24 x^{2} - 82 x + 48 = 0

What are the solutions to the equation 4=x^2+10x ?
The solutions to the equation 4=x^2+10x are x=-5+sqrt(29),x=-5-sqrt(29)
Find the zeros of 4=x^2+10x
The zeros of 4=x^2+10x are x=-5+sqrt(29),x=-5-sqrt(29)