What are the solutions to the equation x^2+10x=39 ?

The solutions to the equation x^2+10x=39 are x=3,x=-13

Find the zeros of x^2+10x=39

The zeros of x^2+10x=39 are x=3,x=-13

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x^{2} + 10 x = 39

x = 3, x = - 13

Solution steps

x^{2} + 10 x = 39

Move

39

to the left side

x^{2} + 10 x - 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 39 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 39) = 16

x_{1, 2} = \frac{- 10 \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 16}{2 \cdot 1}, x_{2} = \frac{- 10 - 16}{2 \cdot 1}

x = \frac{- 10 + 16}{2 \cdot 1} : 3

x = \frac{- 10 - 16}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

x = 3, x = - 13

1 - [\frac{- 5}{100} e^{- 5} - 0.01]

\frac{\frac{π}{2}}{5}

- 2 (2 x - 1) = 3 x

- 4 y^{3} - 116 y^{2} - 400 y

\frac{11 !}{3 !}

What are the solutions to the equation x^2+10x=39 ?
The solutions to the equation x^2+10x=39 are x=3,x=-13
Find the zeros of x^2+10x=39
The zeros of x^2+10x=39 are x=3,x=-13