What are the solutions to the equation 2x^2-14x+9=(x-4)^2 ?

The solutions to the equation 2x^2-14x+9=(x-4)^2 are x=7,x=-1

Find the zeros of 2x^2-14x+9=(x-4)^2

The zeros of 2x^2-14x+9=(x-4)^2 are x=7,x=-1

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2 x^{2} - 14 x + 9 = (x - 4)^{2}

x = 7, x = - 1

Solution steps

2 x^{2} - 14 x + 9 = (x - 4)^{2}

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

2 x^{2} - 14 x + 9 = x^{2} - 8 x + 16

Move

16

to the left side

2 x^{2} - 14 x - 7 = x^{2} - 8 x

Move

8 x

to the left side

2 x^{2} - 6 x - 7 = x^{2}

Move

x^{2}

to the left side

x^{2} - 6 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 7) = 8

x_{1, 2} = \frac{- ( - 6 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 8}{2 \cdot 1}

x = \frac{- ( - 6 ) + 8}{2 \cdot 1} : 7

x = \frac{- ( - 6 ) - 8}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 7, x = - 1

g (- 2) = \frac{- 2 - 4}{- 2 + 10 - 36}

- 2 (x - 5) + 4 x = x - 4

x^{2} + 39 x + 360

\frac{9 - x}{11} = - 4

\frac{4}{3 x - 2} \leq \frac{2}{x + 2}

What are the solutions to the equation 2x^2-14x+9=(x-4)^2 ?
The solutions to the equation 2x^2-14x+9=(x-4)^2 are x=7,x=-1
Find the zeros of 2x^2-14x+9=(x-4)^2
The zeros of 2x^2-14x+9=(x-4)^2 are x=7,x=-1