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

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

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

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

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w^{2} - 4 w = 21

w = 7, w = - 3

Solution steps

w^{2} - 4 w = 21

Move

21

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

w = 7, w = - 3

125 a^{6} - 64

4 - 3 x \geq 0

4 \cdot \frac{1}{5}

∣ 3 - 6 x ∣ > 9

\frac{4}{∣ 2 x - 1 ∣} + 1 = 5

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