What are the solutions to the equation -4=4x^2+3x-6 ?

The solutions to the equation -4=4x^2+3x-6 are x=(-3+sqrt(41))/8 ,x=(-3-sqrt(41))/8

Find the zeros of -4=4x^2+3x-6

The zeros of -4=4x^2+3x-6 are x=(-3+sqrt(41))/8 ,x=(-3-sqrt(41))/8

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- 4 = 4 x^{2} + 3 x - 6

x = \frac{- 3 + 41}{8}, x = \frac{- 3 - 41}{8}

Decimal Notation

x = 0.42539 \dots, x = - 1.17539 \dots

Solution steps

- 4 = 4 x^{2} + 3 x - 6

Switch sides

4 x^{2} + 3 x - 6 = - 4

Move

4

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 4 (- 2) = 41

x_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 41}{2 \cdot 4}, x_{2} = \frac{- 3 - 41}{2 \cdot 4}

x = \frac{- 3 + 41}{2 \cdot 4} : \frac{- 3 + 41}{8}

x = \frac{- 3 - 41}{2 \cdot 4} : \frac{- 3 - 41}{8}

The solutions to the quadratic equation are:

x = \frac{- 3 + 41}{8}, x = \frac{- 3 - 41}{8}

(6 g^{4} h^{5})^{2}

x^{2} - 256 y^{2}

(0 - 1)^{2} + (y + 8)^{2} = 4

\frac{x ^{2} - 4 x + 3}{x ^{2} - 6 x + 8} > 0

x^{2} \leq 2 x

What are the solutions to the equation -4=4x^2+3x-6 ?
The solutions to the equation -4=4x^2+3x-6 are x=(-3+sqrt(41))/8 ,x=(-3-sqrt(41))/8
Find the zeros of -4=4x^2+3x-6
The zeros of -4=4x^2+3x-6 are x=(-3+sqrt(41))/8 ,x=(-3-sqrt(41))/8