What are the solutions to the equation x^2+(2x+20)^2=(2x+22)^2 ?

The solutions to the equation x^2+(2x+20)^2=(2x+22)^2 are x=14,x=-6

Find the zeros of x^2+(2x+20)^2=(2x+22)^2

The zeros of x^2+(2x+20)^2=(2x+22)^2 are x=14,x=-6

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x^{2} + (2 x + 20)^{2} = (2 x + 22)^{2}

x = 14, x = - 6

Solution steps

x^{2} + (2 x + 20)^{2} = (2 x + 22)^{2}

Expand

x^{2} + (2 x + 20)^{2} : 5 x^{2} + 80 x + 400

Expand

(2 x + 22)^{2} : 4 x^{2} + 88 x + 484

5 x^{2} + 80 x + 400 = 4 x^{2} + 88 x + 484

Move

484

to the left side

5 x^{2} + 80 x - 84 = 4 x^{2} + 88 x

Move

88 x

to the left side

5 x^{2} - 8 x - 84 = 4 x^{2}

Move

4 x^{2}

to the left side

x^{2} - 8 x - 84 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 84 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 84) = 20

x_{1, 2} = \frac{- ( - 8 ) \pm 20}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 20}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 20}{2 \cdot 1}

x = \frac{- ( - 8 ) + 20}{2 \cdot 1} : 14

x = \frac{- ( - 8 ) - 20}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 14, x = - 6

\frac{5 - - 121}{1 + - 25}

\frac{10 \cdot 3.2}{17.5}

x^{2} - 2 x + 1 = 0

2 - y = 2

y^{2} + 3 y - 3 = 0

What are the solutions to the equation x^2+(2x+20)^2=(2x+22)^2 ?
The solutions to the equation x^2+(2x+20)^2=(2x+22)^2 are x=14,x=-6
Find the zeros of x^2+(2x+20)^2=(2x+22)^2
The zeros of x^2+(2x+20)^2=(2x+22)^2 are x=14,x=-6