What are the solutions to the equation x^2+28=16x ?

The solutions to the equation x^2+28=16x are x=14,x=2

Find the zeros of x^2+28=16x

The zeros of x^2+28=16x are x=14,x=2

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x^{2} + 28 = 16 x

x = 14, x = 2

Solution steps

x^{2} + 28 = 16 x

Move

16 x

to the left side

x^{2} + 28 - 16 x = 0

Write in the standard form

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

x^{2} - 16 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 28 = 12

x_{1, 2} = \frac{- ( - 16 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 12}{2 \cdot 1}

x = \frac{- ( - 16 ) + 12}{2 \cdot 1} : 14

x = \frac{- ( - 16 ) - 12}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 14, x = 2

7 n^{2} - 6 n - 1 = 0

5 x^{2} + 7 x = x^{2} + 2

5 x^{2} = 1

10 x^{2} + 31 x + 15

\frac{1}{( \frac{2}{3} )}

What are the solutions to the equation x^2+28=16x ?
The solutions to the equation x^2+28=16x are x=14,x=2
Find the zeros of x^2+28=16x
The zeros of x^2+28=16x are x=14,x=2