What is the answer to (x+1)^2-20=(x-3)^2 ?

The answer to (x+1)^2-20=(x-3)^2 is x= 7/2

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(x+1)^2-20=(x-3)^2

(x + 1)^{2} - 20 = (x - 3)^{2}

x = \frac{7}{2}

Decimal Notation

x = 3.5

Solution steps

(x + 1)^{2} - 20 = (x - 3)^{2}

Expand

(x + 1)^{2} - 20 : x^{2} + 2 x - 19

Expand

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

x^{2} + 2 x - 19 = x^{2} - 6 x + 9

Move

19

to the right side

x^{2} + 2 x = x^{2} - 6 x + 28

Subtract

x^{2} - 6 x

from both sides

x^{2} + 2 x - (x^{2} - 6 x) = x^{2} - 6 x + 28 - (x^{2} - 6 x)

Simplify

8 x = 28

Divide both sides by

8

x = \frac{7}{2}

0 = - 3 x + 6

\frac{5 b}{6 a} + \frac{3 b}{10 a ^{2}} + \frac{2}{a b ^{2}}

lo g_{25} (5)

2 p \geq 4

(x - 1) (x - 3) > 0

What is the answer to (x+1)^2-20=(x-3)^2 ?
The answer to (x+1)^2-20=(x-3)^2 is x= 7/2
Solve for x: (x+1)^2-20=(x-3)^2
The solution is x= 7/2