What are the solutions to the equation 10=x^2+3x ?

The solutions to the equation 10=x^2+3x are x=2,x=-5

Find the zeros of 10=x^2+3x

The zeros of 10=x^2+3x are x=2,x=-5

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10 = x^{2} + 3 x

x = 2, x = - 5

Solution steps

10 = x^{2} + 3 x

Switch sides

x^{2} + 3 x = 10

Move

10

to the left side

x^{2} + 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 1}, x_{2} = \frac{- 3 - 7}{2 \cdot 1}

x = \frac{- 3 + 7}{2 \cdot 1} : 2

x = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

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

a, 2 a^{2} b = 2 a^{2} b (1 - 2 a^{3} b^{2})

2 k (k - 5) = 0

- 3 x^{2} + 4 = 0

- x^{2} - 6 x + 7 = 0

What are the solutions to the equation 10=x^2+3x ?
The solutions to the equation 10=x^2+3x are x=2,x=-5
Find the zeros of 10=x^2+3x
The zeros of 10=x^2+3x are x=2,x=-5