What are the solutions to the equation 2(3x-2)^2-2(x-7)(x-5)=-x ?

The solutions to the equation 2(3x-2)^2-2(x-7)(x-5)=-x are x= 31/16 ,x=-2

Find the zeros of 2(3x-2)^2-2(x-7)(x-5)=-x

The zeros of 2(3x-2)^2-2(x-7)(x-5)=-x are x= 31/16 ,x=-2

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2 (3 x - 2)^{2} - 2 (x - 7) (x - 5) = - x

x = \frac{31}{16}, x = - 2

Decimal Notation

x = 1.9375, x = - 2

Solution steps

2 (3 x - 2)^{2} - 2 (x - 7) (x - 5) = - x

Expand

2 (3 x - 2)^{2} - 2 (x - 7) (x - 5) : 16 x^{2} - 62

16 x^{2} - 62 = - x

Move

x

to the left side

16 x^{2} - 62 + x = 0

Write in the standard form

a x^{2} + b x + c = 0

16 x^{2} + x - 62 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 16 ( - 62 )}{2 \cdot 16}

1^{2} - 4 \cdot 16 (- 62) = 63

x_{1, 2} = \frac{- 1 \pm 63}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 1 + 63}{2 \cdot 16}, x_{2} = \frac{- 1 - 63}{2 \cdot 16}

x = \frac{- 1 + 63}{2 \cdot 16} : \frac{31}{16}

x = \frac{- 1 - 63}{2 \cdot 16} : - 2

The solutions to the quadratic equation are:

x = \frac{31}{16}, x = - 2

\frac{x}{2} \geq \frac{25}{x + 1} + 2

t^{2} - 6 t + 8 = 0

\frac{2 x}{3} - \frac{5 x - 7}{6} = \frac{x}{2} + \frac{5}{3}

3 ∣ p + 2 ∣ \geq 6

3 x^{2} - 6 x y - 5 x + 10 y

What are the solutions to the equation 2(3x-2)^2-2(x-7)(x-5)=-x ?
The solutions to the equation 2(3x-2)^2-2(x-7)(x-5)=-x are x= 31/16 ,x=-2
Find the zeros of 2(3x-2)^2-2(x-7)(x-5)=-x
The zeros of 2(3x-2)^2-2(x-7)(x-5)=-x are x= 31/16 ,x=-2