What are the solutions to the equation x^2+14=10x ?

The solutions to the equation x^2+14=10x are x=5+sqrt(11),x=5-sqrt(11)

Find the zeros of x^2+14=10x

The zeros of x^2+14=10x are x=5+sqrt(11),x=5-sqrt(11)

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x^{2} + 14 = 10 x

x = 5 + 11, x = 5 - 11

Decimal Notation

x = 8.31662 \dots, x = 1.68337 \dots

Solution steps

x^{2} + 14 = 10 x

Move

10 x

to the left side

x^{2} + 14 - 10 x = 0

Write in the standard form

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

x^{2} - 10 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 14 = 211

x_{1, 2} = \frac{- ( - 10 ) \pm 2 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 11}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 2 11}{2 \cdot 1}

x = \frac{- ( - 10 ) + 2 11}{2 \cdot 1} : 5 + 11

x = \frac{- ( - 10 ) - 2 11}{2 \cdot 1} : 5 - 11

The solutions to the quadratic equation are:

x = 5 + 11, x = 5 - 11

(m + 1) x^{2} - (m + 8) x + 10 = 0

x^{2} - 12 x + 36 = 144

3 x^{2} - 8 x = 0

x^{2} - \frac{10}{4} = \frac{15}{4}

3 x = \frac{1}{4} x^{2} + 5

What are the solutions to the equation x^2+14=10x ?
The solutions to the equation x^2+14=10x are x=5+sqrt(11),x=5-sqrt(11)
Find the zeros of x^2+14=10x
The zeros of x^2+14=10x are x=5+sqrt(11),x=5-sqrt(11)