What are the solutions to the equation 6=x^2+x ?

The solutions to the equation 6=x^2+x are x=2,x=-3

Find the zeros of 6=x^2+x

The zeros of 6=x^2+x are x=2,x=-3

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6 = x^{2} + x

x = 2, x = - 3

Solution steps

6 = x^{2} + x

Switch sides

x^{2} + x = 6

Move

6

to the left side

x^{2} + x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 \cdot 1}, x_{2} = \frac{- 1 - 5}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2, x = - 3

x^{2} + 12 x = - 27

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

(a + 16)^{2} = 16

s^{2} + 7 s t - 30 t^{2}

4 v^{2} + 15 v + 9

What are the solutions to the equation 6=x^2+x ?
The solutions to the equation 6=x^2+x are x=2,x=-3
Find the zeros of 6=x^2+x
The zeros of 6=x^2+x are x=2,x=-3