What are the solutions to the equation 14=x^2+10x ?

The solutions to the equation 14=x^2+10x are x=-5+sqrt(39),x=-5-sqrt(39)

Find the zeros of 14=x^2+10x

The zeros of 14=x^2+10x are x=-5+sqrt(39),x=-5-sqrt(39)

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14 = x^{2} + 10 x

x = - 5 + 39, x = - 5 - 39

Solution steps

14 = x^{2} + 10 x

Switch sides

x^{2} + 10 x = 14

Move

14

to the left side

x^{2} + 10 x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 14 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 14) = 239

x_{1, 2} = \frac{- 10 \pm 2 39}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 39}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 39}{2 \cdot 1}

x = \frac{- 10 + 2 39}{2 \cdot 1} : - 5 + 39

x = \frac{- 10 - 2 39}{2 \cdot 1} : - 5 - 39

The solutions to the quadratic equation are:

x = - 5 + 39, x = - 5 - 39

5 t^{2} + 7 t - 200 = 0

w^{2} - 64 = 0

5 x^{2} + 30 x - 9 = 0

10 - 10 t = 5 t - 2 t^{2}

z, f = z^{2} - 5 z + 22 f (- 2)

What are the solutions to the equation 14=x^2+10x ?
The solutions to the equation 14=x^2+10x are x=-5+sqrt(39),x=-5-sqrt(39)
Find the zeros of 14=x^2+10x
The zeros of 14=x^2+10x are x=-5+sqrt(39),x=-5-sqrt(39)