What are the solutions to the equation a^2+20a-24=-2 ?

The solutions to the equation a^2+20a-24=-2 are a=-10+sqrt(122),a=-10-sqrt(122)

Find the zeros of a^2+20a-24=-2

The zeros of a^2+20a-24=-2 are a=-10+sqrt(122),a=-10-sqrt(122)

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a^{2} + 20 a - 24 = - 2

a = - 10 + 122, a = - 10 - 122

Decimal Notation

a = 1.04536 \dots, a = - 21.04536 \dots

Solution steps

a^{2} + 20 a - 24 = - 2

Move

2

to the left side

a^{2} + 20 a - 22 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 22 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 22) = 2122

a_{1, 2} = \frac{- 20 \pm 2 122}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 20 + 2 122}{2 \cdot 1}, a_{2} = \frac{- 20 - 2 122}{2 \cdot 1}

a = \frac{- 20 + 2 122}{2 \cdot 1} : - 10 + 122

a = \frac{- 20 - 2 122}{2 \cdot 1} : - 10 - 122

The solutions to the quadratic equation are:

a = - 10 + 122, a = - 10 - 122

0 \leq 2 n + 3 n

(- \frac{5}{6})^{2}

0 = x^{2} + 2 x - 15

5 x^{2} + 7 x \leq 6

8 y^{2} - 16 y

What are the solutions to the equation a^2+20a-24=-2 ?
The solutions to the equation a^2+20a-24=-2 are a=-10+sqrt(122),a=-10-sqrt(122)
Find the zeros of a^2+20a-24=-2
The zeros of a^2+20a-24=-2 are a=-10+sqrt(122),a=-10-sqrt(122)