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

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

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

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

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21 = x^{2} - 4 x

x = 7, x = - 3

Solution steps

21 = x^{2} - 4 x

Switch sides

x^{2} - 4 x = 21

Move

21

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 7, x = - 3

\frac{1}{4} (x + 3)^{2} + 6 = 0

2 y^{2} - 7 y = 7

0.09 x^{2} = 1

665 - 216 x + 12 x^{2} = 0

(x - 5) (x - 8) = 8

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