What are the solutions to the equation -48x=12x^2-45 ?

The solutions to the equation -48x=12x^2-45 are x=(-4+sqrt(31))/2 ,x=-(4+sqrt(31))/2

Find the zeros of -48x=12x^2-45

The zeros of -48x=12x^2-45 are x=(-4+sqrt(31))/2 ,x=-(4+sqrt(31))/2

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- 48 x = 12 x^{2} - 45

x = \frac{- 4 + 31}{2}, x = - \frac{4 + 31}{2}

Decimal

x = 0.78388 \dots, x = - 4.78388 \dots

Solution steps

- 48 x = 12 x^{2} - 45

Switch sides

12 x^{2} - 45 = - 48 x

Move

48 x

to the left side

12 x^{2} - 45 + 48 x = 0

Write in the standard form

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

12 x^{2} + 48 x - 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 12 ( - 45 )}{2 \cdot 12}

4 8^{2} - 4 \cdot 12 (- 45) = 1231

x_{1, 2} = \frac{- 48 \pm 12 31}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 48 + 12 31}{2 \cdot 12}, x_{2} = \frac{- 48 - 12 31}{2 \cdot 12}

x = \frac{- 48 + 12 31}{2 \cdot 12} : \frac{- 4 + 31}{2}

x = \frac{- 48 - 12 31}{2 \cdot 12} : - \frac{4 + 31}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 31}{2}, x = - \frac{4 + 31}{2}

2 0^{2} + 1 4^{2} = z^{2}

110 - p^{2} = p

10 x^{2} + 9 x + 2 = 0

2 W^{2} + 5 W - 63 = 0

x^{2} - 10 x + 16

What are the solutions to the equation -48x=12x^2-45 ?
The solutions to the equation -48x=12x^2-45 are x=(-4+sqrt(31))/2 ,x=-(4+sqrt(31))/2
Find the zeros of -48x=12x^2-45
The zeros of -48x=12x^2-45 are x=(-4+sqrt(31))/2 ,x=-(4+sqrt(31))/2