What are the solutions to the equation -4x^2+x-15=-5x^2 ?

The solutions to the equation -4x^2+x-15=-5x^2 are x=(-1+sqrt(61))/2 ,x=(-1-sqrt(61))/2

Find the zeros of -4x^2+x-15=-5x^2

The zeros of -4x^2+x-15=-5x^2 are x=(-1+sqrt(61))/2 ,x=(-1-sqrt(61))/2

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- 4 x^{2} + x - 15 = - 5 x^{2}

x = \frac{- 1 + 61}{2}, x = \frac{- 1 - 61}{2}

Solution steps

- 4 x^{2} + x - 15 = - 5 x^{2}

Move

5 x^{2}

to the left side

x^{2} + x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 15) = 61

x_{1, 2} = \frac{- 1 \pm 61}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 61}{2 \cdot 1}, x_{2} = \frac{- 1 - 61}{2 \cdot 1}

x = \frac{- 1 + 61}{2 \cdot 1} : \frac{- 1 + 61}{2}

x = \frac{- 1 - 61}{2 \cdot 1} : \frac{- 1 - 61}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 61}{2}, x = \frac{- 1 - 61}{2}

72 x^{2} - 55 x - 150

5 - x \geq 9

- 5 x - 2 \leq - 17

26 = \frac{m}{6} + 5

- 2 x^{2} + 11 x = 0

What are the solutions to the equation -4x^2+x-15=-5x^2 ?
The solutions to the equation -4x^2+x-15=-5x^2 are x=(-1+sqrt(61))/2 ,x=(-1-sqrt(61))/2
Find the zeros of -4x^2+x-15=-5x^2
The zeros of -4x^2+x-15=-5x^2 are x=(-1+sqrt(61))/2 ,x=(-1-sqrt(61))/2