What are the solutions to the equation (x+4)^2-(2x+3)(x-2)=42 ?

The solutions to the equation (x+4)^2-(2x+3)(x-2)=42 are x=4,x=5

Find the zeros of (x+4)^2-(2x+3)(x-2)=42

The zeros of (x+4)^2-(2x+3)(x-2)=42 are x=4,x=5

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(x + 4)^{2} - (2 x + 3) (x - 2) = 42

x = 4, x = 5

Solution steps

(x + 4)^{2} - (2 x + 3) (x - 2) = 42

Expand

(x + 4)^{2} - (2 x + 3) (x - 2) : - x^{2} + 9 x + 22

- x^{2} + 9 x + 22 = 42

Move

42

to the left side

- x^{2} + 9 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 1 ) ( - 20 )}{2 ( - 1 )}

9^{2} - 4 (- 1) (- 20) = 1

x_{1, 2} = \frac{- 9 \pm 1}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 9 + 1}{2 ( - 1 )}, x_{2} = \frac{- 9 - 1}{2 ( - 1 )}

x = \frac{- 9 + 1}{2 ( - 1 )} : 4

x = \frac{- 9 - 1}{2 ( - 1 )} : 5

The solutions to the quadratic equation are:

x = 4, x = 5

6 (x + 1) - (4 - x) (x + 1) = 6 (4 - x)

y, 4 x^{3} - 2 x = 2 y^{2} + 7

λ^{2} - 4 = 0

3 x^{2} - 8 x + 5 = 0

4 x (x - 6) = - 9

What are the solutions to the equation (x+4)^2-(2x+3)(x-2)=42 ?
The solutions to the equation (x+4)^2-(2x+3)(x-2)=42 are x=4,x=5
Find the zeros of (x+4)^2-(2x+3)(x-2)=42
The zeros of (x+4)^2-(2x+3)(x-2)=42 are x=4,x=5