What are the solutions to the equation x^2+36=14x ?

The solutions to the equation x^2+36=14x are x=7+sqrt(13),x=7-sqrt(13)

Find the zeros of x^2+36=14x

The zeros of x^2+36=14x are x=7+sqrt(13),x=7-sqrt(13)

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x^{2} + 36 = 14 x

x = 7 + 13, x = 7 - 13

Decimal Notation

x = 10.60555 \dots, x = 3.39444 \dots

Solution steps

x^{2} + 36 = 14 x

Move

14 x

to the left side

x^{2} + 36 - 14 x = 0

Write in the standard form

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

x^{2} - 14 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 36 = 213

x_{1, 2} = \frac{- ( - 14 ) \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 13}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 2 13}{2 \cdot 1}

x = \frac{- ( - 14 ) + 2 13}{2 \cdot 1} : 7 + 13

x = \frac{- ( - 14 ) - 2 13}{2 \cdot 1} : 7 - 13

The solutions to the quadratic equation are:

x = 7 + 13, x = 7 - 13

\frac{1}{4} + \frac{1}{6} + \frac{1}{12}

\frac{x ^{7} y ^{4}}{x ^{9} y ^{5}}

\frac{1}{x} + \frac{1}{x + 1} < \frac{2}{x + 5}

\frac{\frac{0.51}{204.22}}{0.0136}

x^{2} - 5 = 0

What are the solutions to the equation x^2+36=14x ?
The solutions to the equation x^2+36=14x are x=7+sqrt(13),x=7-sqrt(13)
Find the zeros of x^2+36=14x
The zeros of x^2+36=14x are x=7+sqrt(13),x=7-sqrt(13)