What are the solutions to the equation 3x+7-5x^2=-4 ?

The solutions to the equation 3x+7-5x^2=-4 are x=-(-3+sqrt(229))/(10),x=(3+sqrt(229))/(10)

Find the zeros of 3x+7-5x^2=-4

The zeros of 3x+7-5x^2=-4 are x=-(-3+sqrt(229))/(10),x=(3+sqrt(229))/(10)

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3 x + 7 - 5 x^{2} = - 4

x = - \frac{- 3 + 229}{10}, x = \frac{3 + 229}{10}

Decimal Notation

x = - 1.21327 \dots, x = 1.81327 \dots

Solution steps

3 x + 7 - 5 x^{2} = - 4

Move

4

to the left side

- 5 x^{2} + 3 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 5 ) \cdot 11}{2 ( - 5 )}

3^{2} - 4 (- 5) \cdot 11 = 229

x_{1, 2} = \frac{- 3 \pm 229}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 3 + 229}{2 ( - 5 )}, x_{2} = \frac{- 3 - 229}{2 ( - 5 )}

x = \frac{- 3 + 229}{2 ( - 5 )} : - \frac{- 3 + 229}{10}

x = \frac{- 3 - 229}{2 ( - 5 )} : \frac{3 + 229}{10}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 229}{10}, x = \frac{3 + 229}{10}

- 3 x^{2} + 5 x + 8 = 0

x^{2} + 4 x + 29

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

x^{2} - 8 x = 0

q^{2} = - 9

What are the solutions to the equation 3x+7-5x^2=-4 ?
The solutions to the equation 3x+7-5x^2=-4 are x=-(-3+sqrt(229))/(10),x=(3+sqrt(229))/(10)
Find the zeros of 3x+7-5x^2=-4
The zeros of 3x+7-5x^2=-4 are x=-(-3+sqrt(229))/(10),x=(3+sqrt(229))/(10)