What are the solutions to the equation x^2+71=18x-9 ?

The solutions to the equation x^2+71=18x-9 are x=10,x=8

Find the zeros of x^2+71=18x-9

The zeros of x^2+71=18x-9 are x=10,x=8

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x^{2} + 71 = 18 x - 9

x = 10, x = 8

Solution steps

x^{2} + 71 = 18 x - 9

Move

9

to the left side

x^{2} + 80 = 18 x

Move

18 x

to the left side

x^{2} + 80 - 18 x = 0

Write in the standard form

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

x^{2} - 18 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 80 = 2

x_{1, 2} = \frac{- ( - 18 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 2}{2 \cdot 1}

x = \frac{- ( - 18 ) + 2}{2 \cdot 1} : 10

x = \frac{- ( - 18 ) - 2}{2 \cdot 1} : 8

The solutions to the quadratic equation are:

x = 10, x = 8

m^{2} - 3 m + 5 = 0

(y + 7)^{2} = 2 y^{2} + 17 + 39

x, y^{2} - x^{2} = c

3 x^{2} - 7 x + 2 = 0

x^{2} + \frac{18}{5} x + \frac{81}{25} = 0

What are the solutions to the equation x^2+71=18x-9 ?
The solutions to the equation x^2+71=18x-9 are x=10,x=8
Find the zeros of x^2+71=18x-9
The zeros of x^2+71=18x-9 are x=10,x=8