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

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

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

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

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x^{2} - 4 x = 21

x = 7, x = - 3

Solution steps

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

- 19 \cdot - 19

37

3 y

\frac{50}{32}

x, 1 0^{x} = 15

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