What are the solutions to the equation -3x-8=-4x^2 ?

The solutions to the equation -3x-8=-4x^2 are x=-(-3+sqrt(137))/8 ,x=(3+sqrt(137))/8

Find the zeros of -3x-8=-4x^2

The zeros of -3x-8=-4x^2 are x=-(-3+sqrt(137))/8 ,x=(3+sqrt(137))/8

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- 3 x - 8 = - 4 x^{2}

x = - \frac{- 3 + 137}{8}, x = \frac{3 + 137}{8}

Solution steps

- 3 x - 8 = - 4 x^{2}

Switch sides

- 4 x^{2} = - 3 x - 8

Move

8

to the left side

- 4 x^{2} + 8 = - 3 x

Move

3 x

to the left side

- 4 x^{2} + 8 + 3 x = 0

Write in the standard form

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

- 4 x^{2} + 3 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 4 ) \cdot 8}{2 ( - 4 )}

3^{2} - 4 (- 4) \cdot 8 = 137

x_{1, 2} = \frac{- 3 \pm 137}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 3 + 137}{2 ( - 4 )}, x_{2} = \frac{- 3 - 137}{2 ( - 4 )}

x = \frac{- 3 + 137}{2 ( - 4 )} : - \frac{- 3 + 137}{8}

x = \frac{- 3 - 137}{2 ( - 4 )} : \frac{3 + 137}{8}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 137}{8}, x = \frac{3 + 137}{8}

7 x^{2} + 6 x + 2 = 0

- 0.04

7 x^{2} + 2 x - 5

(5 x - 2)^{2} \leq 0

4 (2 x - 9) = 4

What are the solutions to the equation -3x-8=-4x^2 ?
The solutions to the equation -3x-8=-4x^2 are x=-(-3+sqrt(137))/8 ,x=(3+sqrt(137))/8
Find the zeros of -3x-8=-4x^2
The zeros of -3x-8=-4x^2 are x=-(-3+sqrt(137))/8 ,x=(3+sqrt(137))/8