What are the solutions to the equation -(x-4)(2x+3)=2(4x+5) ?

The solutions to the equation -(x-4)(2x+3)=2(4x+5) are x=-2,x= 1/2

Find the zeros of -(x-4)(2x+3)=2(4x+5)

The zeros of -(x-4)(2x+3)=2(4x+5) are x=-2,x= 1/2

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- (x - 4) (2 x + 3) = 2 (4 x + 5)

x = - 2, x = \frac{1}{2}

Decimal Notation

x = - 2, x = 0.5

Solution steps

- (x - 4) (2 x + 3) = 2 (4 x + 5)

Expand

- (x - 4) (2 x + 3) : - 2 x^{2} + 5 x + 12

Expand

2 (4 x + 5) : 8 x + 10

- 2 x^{2} + 5 x + 12 = 8 x + 10

Move

10

to the left side

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

Move

8 x

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

x = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

The solutions to the quadratic equation are:

x = - 2, x = \frac{1}{2}

32 = 16 + 32 t - 16 t^{2}

u 7.9 u - 2 - 3.2 u - 10 = 2.8 + u

5 x - 3 = 2 x^{2}

x^{2} + 12 x = 0

x, - 6 x y - x^{2} + 3 x = 0

What are the solutions to the equation -(x-4)(2x+3)=2(4x+5) ?
The solutions to the equation -(x-4)(2x+3)=2(4x+5) are x=-2,x= 1/2
Find the zeros of -(x-4)(2x+3)=2(4x+5)
The zeros of -(x-4)(2x+3)=2(4x+5) are x=-2,x= 1/2