What are the solutions to the equation x^2+x=42 ?

The solutions to the equation x^2+x=42 are x=6,x=-7

Find the zeros of x^2+x=42

The zeros of x^2+x=42 are x=6,x=-7

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x^{2} + x = 42

x = 6, x = - 7

Solution steps

x^{2} + x = 42

Move

42

to the left side

x^{2} + x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 1}, x_{2} = \frac{- 1 - 13}{2 \cdot 1}

x = \frac{- 1 + 13}{2 \cdot 1} : 6

x = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 6, x = - 7

4 p^{2} - 4 p - 15 = 0

16 x^{2} - 169 y^{2}

9 t^{3} - 90 t^{2} + 144 t

(\frac{1}{3} - i \frac{2 5}{3} - \frac{1}{3})^{2}

4 a - 3 + 2 a = 7 a - 2

What are the solutions to the equation x^2+x=42 ?
The solutions to the equation x^2+x=42 are x=6,x=-7
Find the zeros of x^2+x=42
The zeros of x^2+x=42 are x=6,x=-7