What are the solutions to the equation x^2+98=20x ?

The solutions to the equation x^2+98=20x are x=10+sqrt(2),x=10-sqrt(2)

Find the zeros of x^2+98=20x

The zeros of x^2+98=20x are x=10+sqrt(2),x=10-sqrt(2)

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x^{2} + 98 = 20 x

x = 10 + 2, x = 10 - 2

Solution steps

x^{2} + 98 = 20 x

Move

20 x

to the left side

x^{2} + 98 - 20 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 20 x + 98 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 98}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 98 = 22

x_{1, 2} = \frac{- ( - 20 ) \pm 2 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 2 2}{2 \cdot 1}, x_{2} = \frac{- ( - 20 ) - 2 2}{2 \cdot 1}

x = \frac{- ( - 20 ) + 2 2}{2 \cdot 1} : 10 + 2

x = \frac{- ( - 20 ) - 2 2}{2 \cdot 1} : 10 - 2

The solutions to the quadratic equation are:

x = 10 + 2, x = 10 - 2

\frac{1}{4} x - 2 = 3

x^{2} + \frac{x}{5} = 404

62 5^{\frac{5}{4}}

- x^{2} - 3 x + 18 \leq 0

(6 + 7 i) (3 - 7 i)

What are the solutions to the equation x^2+98=20x ?
The solutions to the equation x^2+98=20x are x=10+sqrt(2),x=10-sqrt(2)
Find the zeros of x^2+98=20x
The zeros of x^2+98=20x are x=10+sqrt(2),x=10-sqrt(2)