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

The solutions to the equation x-x^2+3=4x^2+2 are x=-(-1+sqrt(21))/(10),x=(1+sqrt(21))/(10)

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

The zeros of x-x^2+3=4x^2+2 are x=-(-1+sqrt(21))/(10),x=(1+sqrt(21))/(10)

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

x = - \frac{- 1 + 21}{10}, x = \frac{1 + 21}{10}

Decimal Notation

x = - 0.35825 \dots, x = 0.55825 \dots

Solution steps

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

Move

2

to the left side

- x^{2} + x + 1 = 4 x^{2}

Move

4 x^{2}

to the left side

- 5 x^{2} + x + 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 (- 5) \cdot 1 = 21

x_{1, 2} = \frac{- 1 \pm 21}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 1 + 21}{2 ( - 5 )}, x_{2} = \frac{- 1 - 21}{2 ( - 5 )}

x = \frac{- 1 + 21}{2 ( - 5 )} : - \frac{- 1 + 21}{10}

x = \frac{- 1 - 21}{2 ( - 5 )} : \frac{1 + 21}{10}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 21}{10}, x = \frac{1 + 21}{10}

2 x (5 x - 3) = 0

y, h = 16 - 3 y^{2}

x^{2} + 10 x = 5

0 = x^{2} - 12 x + 1

4 x^{2} - 16 x + 22

What are the solutions to the equation x-x^2+3=4x^2+2 ?
The solutions to the equation x-x^2+3=4x^2+2 are x=-(-1+sqrt(21))/(10),x=(1+sqrt(21))/(10)
Find the zeros of x-x^2+3=4x^2+2
The zeros of x-x^2+3=4x^2+2 are x=-(-1+sqrt(21))/(10),x=(1+sqrt(21))/(10)