What are the solutions to the equation x^2+x-156=0 ?

The solutions to the equation x^2+x-156=0 are x=12,x=-13

Find the zeros of x^2+x-156=0

The zeros of x^2+x-156=0 are x=12,x=-13

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x^{2} + x - 156 = 0

x = 12, x = - 13

Solution steps

x^{2} + x - 156 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 156 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 156) = 25

x_{1, 2} = \frac{- 1 \pm 25}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 25}{2 \cdot 1}, x_{2} = \frac{- 1 - 25}{2 \cdot 1}

x = \frac{- 1 + 25}{2 \cdot 1} : 12

x = \frac{- 1 - 25}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

x = 12, x = - 13

lo g_{10} (0.05)

(\frac{2}{5})^{2} + \frac{1}{2} - (\frac{1}{3} - \frac{3}{4}) + (- \frac{1}{2})^{2}

f = 2 (- 5)^{2} - 3 (- 5)

2 (x - 7) - 10 = 12 - 4 x

\frac{x}{9} + 6 < 4

What are the solutions to the equation x^2+x-156=0 ?
The solutions to the equation x^2+x-156=0 are x=12,x=-13
Find the zeros of x^2+x-156=0
The zeros of x^2+x-156=0 are x=12,x=-13