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

The solutions to the equation 2/3 (x-4)(x+5)=1 are x=(-1+sqrt(87))/2 ,x=-(1+sqrt(87))/2

Find the zeros of 2/3 (x-4)(x+5)=1

The zeros of 2/3 (x-4)(x+5)=1 are x=(-1+sqrt(87))/2 ,x=-(1+sqrt(87))/2

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\frac{2}{3} (x - 4) (x + 5) = 1

x = \frac{- 1 + 87}{2}, x = - \frac{1 + 87}{2}

Decimal Notation

x = 4.16368 \dots, x = - 5.16368 \dots

Solution steps

\frac{2}{3} (x - 4) (x + 5) = 1

Multiply both sides by

3

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

Expand

2 (x - 4) (x + 5) : 2 x^{2} + 2 x - 40

2 x^{2} + 2 x - 40 = 3

Move

3

to the left side

2 x^{2} + 2 x - 43 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 43 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 43) = 287

x_{1, 2} = \frac{- 2 \pm 2 87}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 87}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 87}{2 \cdot 2}

x = \frac{- 2 + 2 87}{2 \cdot 2} : \frac{- 1 + 87}{2}

x = \frac{- 2 - 2 87}{2 \cdot 2} : - \frac{1 + 87}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 87}{2}, x = - \frac{1 + 87}{2}

y, x^{2} y + x^{3} y^{2} = 2

t, d (t) = - 9.81 t^{2} + 10 t m

4 z^{2} - r^{2} = 4

- 9 = - 23 + 7 x^{2}

- x^{2} + 5 x + 1 = 0

What are the solutions to the equation 2/3 (x-4)(x+5)=1 ?
The solutions to the equation 2/3 (x-4)(x+5)=1 are x=(-1+sqrt(87))/2 ,x=-(1+sqrt(87))/2
Find the zeros of 2/3 (x-4)(x+5)=1
The zeros of 2/3 (x-4)(x+5)=1 are x=(-1+sqrt(87))/2 ,x=-(1+sqrt(87))/2