What are the solutions to the equation x^2-42x+n=0 ?

The solutions to the equation x^2-42x+n=0 are x=21+sqrt(-n+441),x=21-sqrt(-n+441)

Find the zeros of x^2-42x+n=0

The zeros of x^2-42x+n=0 are x=21+sqrt(-n+441),x=21-sqrt(-n+441)

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x^{2} - 42 x + n = 0

x = 21 + - n + 441, x = 21 - - n + 441

Solution steps

x^{2} - 42 x + n = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 1 \cdot n}{2 \cdot 1}

Simplify

(- 42)^{2} - 4 \cdot 1 \cdot n : 2 441 - n

x_{1, 2} = \frac{- ( - 42 ) \pm 2 441 - n}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 2 441 - n}{2 \cdot 1}, x_{2} = \frac{- ( - 42 ) - 2 441 - n}{2 \cdot 1}

x = \frac{- ( - 42 ) + 2 441 - n}{2 \cdot 1} : 21 + - n + 441

x = \frac{- ( - 42 ) - 2 441 - n}{2 \cdot 1} : 21 - - n + 441

The solutions to the quadratic equation are:

x = 21 + - n + 441, x = 21 - - n + 441

\frac{1}{2} x^{2} - \frac{1}{4} x = 0

1 - (\frac{( 1 - 5 )}{2} x)^{2} - x^{2} = 0

- \frac{3}{2} x^{2} - 12 x + 54 = 0

- 3 x^{2} - 10 x + 16 = 0

8 x^{2} + 22 x + 15 = 0

What are the solutions to the equation x^2-42x+n=0 ?
The solutions to the equation x^2-42x+n=0 are x=21+sqrt(-n+441),x=21-sqrt(-n+441)
Find the zeros of x^2-42x+n=0
The zeros of x^2-42x+n=0 are x=21+sqrt(-n+441),x=21-sqrt(-n+441)