What are the solutions to the equation (x)(x+1)=240 ?

The solutions to the equation (x)(x+1)=240 are x=15,x=-16

Find the zeros of (x)(x+1)=240

The zeros of (x)(x+1)=240 are x=15,x=-16

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(x) (x + 1) = 240

x = 15, x = - 16

Solution steps

(x) (x + 1) = 240

Expand

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

x^{2} + x = 240

Move

240

to the left side

x^{2} + x - 240 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 240 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 240) = 31

x_{1, 2} = \frac{- 1 \pm 31}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 31}{2 \cdot 1}, x_{2} = \frac{- 1 - 31}{2 \cdot 1}

x = \frac{- 1 + 31}{2 \cdot 1} : 15

x = \frac{- 1 - 31}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

x = 15, x = - 16

2 x^{2} - x - 11 = - 3

5 x^{2} + 5 x - \frac{3}{4} = 0

3 x^{2} + 12 x = 7 x + 2

(x - 1)^{2}

s^{2} - 6 s + 25

What are the solutions to the equation (x)(x+1)=240 ?
The solutions to the equation (x)(x+1)=240 are x=15,x=-16
Find the zeros of (x)(x+1)=240
The zeros of (x)(x+1)=240 are x=15,x=-16