What are the solutions to the equation (x+2)^2-(2x-3)^2=-80 ?

The solutions to the equation (x+2)^2-(2x-3)^2=-80 are x=-3,x= 25/3

Find the zeros of (x+2)^2-(2x-3)^2=-80

The zeros of (x+2)^2-(2x-3)^2=-80 are x=-3,x= 25/3

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(x + 2)^{2} - (2 x - 3)^{2} = - 80

x = - 3, x = \frac{25}{3}

Decimal Notation

x = - 3, x = 8.33333 \dots

Solution steps

(x + 2)^{2} - (2 x - 3)^{2} = - 80

Expand

(x + 2)^{2} - (2 x - 3)^{2} : - 3 x^{2} + 16 x - 5

- 3 x^{2} + 16 x - 5 = - 80

Move

80

to the left side

- 3 x^{2} + 16 x + 75 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 3 ) \cdot 75}{2 ( - 3 )}

1 6^{2} - 4 (- 3) \cdot 75 = 34

x_{1, 2} = \frac{- 16 \pm 34}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 16 + 34}{2 ( - 3 )}, x_{2} = \frac{- 16 - 34}{2 ( - 3 )}

x = \frac{- 16 + 34}{2 ( - 3 )} : - 3

x = \frac{- 16 - 34}{2 ( - 3 )} : \frac{25}{3}

The solutions to the quadratic equation are:

x = - 3, x = \frac{25}{3}

\frac{1 5 ^{2}}{4 0 ^{2}} + \frac{y ^{2}}{3 0 ^{2}} = 1

8 x^{2} - x = 0

0 = x^{2} + x + 2

- 2 x^{2} = 18

2 m^{2} - 7 m = 0

What are the solutions to the equation (x+2)^2-(2x-3)^2=-80 ?
The solutions to the equation (x+2)^2-(2x-3)^2=-80 are x=-3,x= 25/3
Find the zeros of (x+2)^2-(2x-3)^2=-80
The zeros of (x+2)^2-(2x-3)^2=-80 are x=-3,x= 25/3