What are the solutions to the equation (x)(x+1)=156 ?

The solutions to the equation (x)(x+1)=156 are x=12,x=-13

Find the zeros of (x)(x+1)=156

The zeros of (x)(x+1)=156 are x=12,x=-13

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(x) (x + 1) = 156

x = 12, x = - 13

Solution steps

(x) (x + 1) = 156

Expand

(x) (x + 1) : x^{2} + x

x^{2} + x = 156

Move

156

to the left side

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

x^{2} - 10 x = 50

12 x^{2} + 8 x + 1 = 0

x, x^{2} + 45 k^{2} = 18 k x

x, 2 x^{2} + 12 x + 4 = 0

m^{2} + 40 m + 4000 = 0

What are the solutions to the equation (x)(x+1)=156 ?
The solutions to the equation (x)(x+1)=156 are x=12,x=-13
Find the zeros of (x)(x+1)=156
The zeros of (x)(x+1)=156 are x=12,x=-13