What are the solutions to the equation a^2-4a=21 ?

The solutions to the equation a^2-4a=21 are a=7,a=-3

Find the zeros of a^2-4a=21

The zeros of a^2-4a=21 are a=7,a=-3

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a^{2} - 4 a = 21

a = 7, a = - 3

Solution steps

a^{2} - 4 a = 21

Move

21

to the left side

a^{2} - 4 a - 21 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

a_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, a_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

a = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

a = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

a = 7, a = - 3

L (p) = - 4000 p^{2} + 66000 p - 140000

3 x - 2 = 2 (x - 3) (2 x - 2)

2 x + \frac{x ^{2}}{2} = 16

x^{2} + 2 x = 20

x^{2} + 2 x = - 8

What are the solutions to the equation a^2-4a=21 ?
The solutions to the equation a^2-4a=21 are a=7,a=-3
Find the zeros of a^2-4a=21
The zeros of a^2-4a=21 are a=7,a=-3