What are the solutions to the equation a\times a+a=240 ?

The solutions to the equation a\times a+a=240 are a=15,a=-16

Find the zeros of a\times a+a=240

The zeros of a\times a+a=240 are a=15,a=-16

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a \cdot a + a = 240

a = 15, a = - 16

Solution steps

aa + a = 240

Expand

aa + a : a^{2} + a

a^{2} + a = 240

Move

240

to the left side

a^{2} + a - 240 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

a = 15, a = - 16

x^{2} - 14 x - 20 = 0

2 x^{2} - 3 = 69

7 x^{2} - 4 x - 20 = 0

x^{2} - 3 x - 54 = 0

- x^{2} + 12 x + 20 = 0

What are the solutions to the equation a\times a+a=240 ?
The solutions to the equation a\times a+a=240 are a=15,a=-16
Find the zeros of a\times a+a=240
The zeros of a\times a+a=240 are a=15,a=-16