What are the solutions to the equation (x)(x+1)=2970 ?

The solutions to the equation (x)(x+1)=2970 are x=54,x=-55

Find the zeros of (x)(x+1)=2970

The zeros of (x)(x+1)=2970 are x=54,x=-55

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(x) (x + 1) = 2970

x = 54, x = - 55

Solution steps

(x) (x + 1) = 2970

Expand

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

x^{2} + x = 2970

Move

2970

to the left side

x^{2} + x - 2970 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2970 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 2970) = 109

x_{1, 2} = \frac{- 1 \pm 109}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 109}{2 \cdot 1}, x_{2} = \frac{- 1 - 109}{2 \cdot 1}

x = \frac{- 1 + 109}{2 \cdot 1} : 54

x = \frac{- 1 - 109}{2 \cdot 1} : - 55

The solutions to the quadratic equation are:

x = 54, x = - 55

0 = - 16 t^{2} + 16 t + 32

[\frac{( x + 3 )}{( x + 8 )}] \geq 6

19 - 3 x \leq - (1 + 7 x)

- (a + 1) - 4 a \leq 2 a - 8

- \frac{6}{1}

What are the solutions to the equation (x)(x+1)=2970 ?
The solutions to the equation (x)(x+1)=2970 are x=54,x=-55
Find the zeros of (x)(x+1)=2970
The zeros of (x)(x+1)=2970 are x=54,x=-55