What are the solutions to the equation x^2+x=12 ?

The solutions to the equation x^2+x=12 are x=3,x=-4

Find the zeros of x^2+x=12

The zeros of x^2+x=12 are x=3,x=-4

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x^{2} + x = 12

x = 3, x = - 4

Solution steps

x^{2} + x = 12

Move

12

to the left side

x^{2} + x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

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

Separate the solutions

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

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

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

\frac{5 x + 7}{2} - \frac{x + 4}{5} = 1 - (\frac{3 - x}{4})

(x - 3) (x + 5) = (x + 5)^{2}

72 t^{2} + t - 1

6 u + 5 = 8 u + 11

a^{2} + a^{3}

What are the solutions to the equation x^2+x=12 ?
The solutions to the equation x^2+x=12 are x=3,x=-4
Find the zeros of x^2+x=12
The zeros of x^2+x=12 are x=3,x=-4