What are the solutions to the equation (x-4)^2=2x^2-15+22 ?

The solutions to the equation (x-4)^2=2x^2-15+22 are x=-9,x=1

Find the zeros of (x-4)^2=2x^2-15+22

The zeros of (x-4)^2=2x^2-15+22 are x=-9,x=1

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(x - 4)^{2} = 2 x^{2} - 15 + 22

x = - 9, x = 1

Solution steps

(x - 4)^{2} = 2 x^{2} - 15 + 22

Expand

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

x^{2} - 8 x + 16 = 2 x^{2} + 7

Move

7

to the left side

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

Move

2 x^{2}

to the left side

- x^{2} - 8 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) \cdot 9}{2 ( - 1 )}

(- 8)^{2} - 4 (- 1) \cdot 9 = 10

x_{1, 2} = \frac{- ( - 8 ) \pm 10}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 10}{2 ( - 1 )}, x_{2} = \frac{- ( - 8 ) - 10}{2 ( - 1 )}

x = \frac{- ( - 8 ) + 10}{2 ( - 1 )} : - 9

x = \frac{- ( - 8 ) - 10}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 9, x = 1

- 5 ∣ - 3 x + 2 ∣ - 2 = - 2

867

\frac{5}{2} (\frac{1}{4})

\frac{5 + m}{5} - \frac{m + 4}{3} \leq - \frac{m}{15}

9 - 18 \div 3^{2}

What are the solutions to the equation (x-4)^2=2x^2-15+22 ?
The solutions to the equation (x-4)^2=2x^2-15+22 are x=-9,x=1
Find the zeros of (x-4)^2=2x^2-15+22
The zeros of (x-4)^2=2x^2-15+22 are x=-9,x=1