What are the solutions to the equation 2x(x+4)=(x-3)^2 ?

The solutions to the equation 2x(x+4)=(x-3)^2 are x=-7-sqrt(58),x=sqrt(58)-7

Find the zeros of 2x(x+4)=(x-3)^2

The zeros of 2x(x+4)=(x-3)^2 are x=-7-sqrt(58),x=sqrt(58)-7

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2 x (x + 4) = (x - 3)^{2}

x = - 7 - 58, x = 58 - 7

Decimal Notation

x = - 14.61577 \dots, x = 0.61577 \dots

Solution steps

2 x (x + 4) = (x - 3)^{2}

Expand

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

Expand

(x - 3)^{2} : x^{2} - 6 x + 9

2 x^{2} + 8 x = x^{2} - 6 x + 9

Switch sides

x^{2} - 6 x + 9 = 2 x^{2} + 8 x

Move

8 x

to the left side

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

Move

2 x^{2}

to the left side

- x^{2} - 14 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 ( - 1 ) \cdot 9}{2 ( - 1 )}

(- 14)^{2} - 4 (- 1) \cdot 9 = 258

x_{1, 2} = \frac{- ( - 14 ) \pm 2 58}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 58}{2 ( - 1 )}, x_{2} = \frac{- ( - 14 ) - 2 58}{2 ( - 1 )}

x = \frac{- ( - 14 ) + 2 58}{2 ( - 1 )} : - 7 - 58

x = \frac{- ( - 14 ) - 2 58}{2 ( - 1 )} : 58 - 7

The solutions to the quadratic equation are:

x = - 7 - 58, x = 58 - 7

x (x + 3) = 16

y, 5 x^{3} - x y^{2} + 3 y = C

7 x^{2} - 10 x = 0

2 x^{2} + 4 x + 11

m^{2} + 11 m + 3 = - 4 m^{2}

What are the solutions to the equation 2x(x+4)=(x-3)^2 ?
The solutions to the equation 2x(x+4)=(x-3)^2 are x=-7-sqrt(58),x=sqrt(58)-7
Find the zeros of 2x(x+4)=(x-3)^2
The zeros of 2x(x+4)=(x-3)^2 are x=-7-sqrt(58),x=sqrt(58)-7