What are the solutions to the equation 6(x^2-1)=-5x ?

The solutions to the equation 6(x^2-1)=-5x are x= 2/3 ,x=-3/2

Find the zeros of 6(x^2-1)=-5x

The zeros of 6(x^2-1)=-5x are x= 2/3 ,x=-3/2

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6 (x^{2} - 1) = - 5 x

x = \frac{2}{3}, x = - \frac{3}{2}

Decimal Notation

x = 0.66666 \dots, x = - 1.5

Solution steps

6 (x^{2} - 1) = - 5 x

Expand

6 (x^{2} - 1) : 6 x^{2} - 6

6 x^{2} - 6 = - 5 x

Move

5 x

to the left side

6 x^{2} - 6 + 5 x = 0

Write in the standard form

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

6 x^{2} + 5 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 ( - 6 )}{2 \cdot 6}

5^{2} - 4 \cdot 6 (- 6) = 13

x_{1, 2} = \frac{- 5 \pm 13}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 5 + 13}{2 \cdot 6}, x_{2} = \frac{- 5 - 13}{2 \cdot 6}

x = \frac{- 5 + 13}{2 \cdot 6} : \frac{2}{3}

x = \frac{- 5 - 13}{2 \cdot 6} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = - \frac{3}{2}

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

36 x^{2} + 48 x + 16 = 12

9 x^{2} - 42 x = - 52

2 x - 3 = x^{2}

t, h = \frac{1}{2} g t^{2}

What are the solutions to the equation 6(x^2-1)=-5x ?
The solutions to the equation 6(x^2-1)=-5x are x= 2/3 ,x=-3/2
Find the zeros of 6(x^2-1)=-5x
The zeros of 6(x^2-1)=-5x are x= 2/3 ,x=-3/2