What are the solutions to the equation x^2-(2x+1)^2=-21 ?

The solutions to the equation x^2-(2x+1)^2=-21 are x=-10/3 ,x=2

Find the zeros of x^2-(2x+1)^2=-21

The zeros of x^2-(2x+1)^2=-21 are x=-10/3 ,x=2

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x^{2} - (2 x + 1)^{2} = - 21

x = - \frac{10}{3}, x = 2

Decimal

x = - 3.33333 \dots, x = 2

Solution steps

x^{2} - (2 x + 1)^{2} = - 21

Expand

x^{2} - (2 x + 1)^{2} : - 3 x^{2} - 4 x - 1

- 3 x^{2} - 4 x - 1 = - 21

Move

21

to the left side

- 3 x^{2} - 4 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 3 ) \cdot 20}{2 ( - 3 )}

(- 4)^{2} - 4 (- 3) \cdot 20 = 16

x_{1, 2} = \frac{- ( - 4 ) \pm 16}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 16}{2 ( - 3 )}, x_{2} = \frac{- ( - 4 ) - 16}{2 ( - 3 )}

x = \frac{- ( - 4 ) + 16}{2 ( - 3 )} : - \frac{10}{3}

x = \frac{- ( - 4 ) - 16}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

x = - \frac{10}{3}, x = 2

0 = - 0.04 x^{2} + 0.8 x + 9

4 x^{2} - 15 x + 14 = 0

x^{2} - 10 = 4 x + 11

47 y^{2} = - 15 y

x^{2} - 10 x + 45 = 0

What are the solutions to the equation x^2-(2x+1)^2=-21 ?
The solutions to the equation x^2-(2x+1)^2=-21 are x=-10/3 ,x=2
Find the zeros of x^2-(2x+1)^2=-21
The zeros of x^2-(2x+1)^2=-21 are x=-10/3 ,x=2