What are the solutions to the equation a^2-22a-168=0 ?

The solutions to the equation a^2-22a-168=0 are a=28,a=-6

Find the zeros of a^2-22a-168=0

The zeros of a^2-22a-168=0 are a=28,a=-6

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a^{2} - 22 a - 168 = 0

a = 28, a = - 6

Solution steps

a^{2} - 22 a - 168 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot ( - 168 )}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot (- 168) = 34

a_{1, 2} = \frac{- ( - 22 ) \pm 34}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 22 ) + 34}{2 \cdot 1}, a_{2} = \frac{- ( - 22 ) - 34}{2 \cdot 1}

a = \frac{- ( - 22 ) + 34}{2 \cdot 1} : 28

a = \frac{- ( - 22 ) - 34}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

a = 28, a = - 6

2 x^{2} - x = - 4

3 x - 7 - 2 x = - 8 x - 57 - x

\frac{( x ^{2} + 1 )}{2} \geq 3

∣ 2 - 0.2 x ∣ = 2

∣ 5 w - 15 ∣ > 5

What are the solutions to the equation a^2-22a-168=0 ?
The solutions to the equation a^2-22a-168=0 are a=28,a=-6
Find the zeros of a^2-22a-168=0
The zeros of a^2-22a-168=0 are a=28,a=-6