What are the solutions to the equation 8x+2+5x^2=4 ?

The solutions to the equation 8x+2+5x^2=4 are x=(-4+sqrt(26))/5 ,x=-(4+sqrt(26))/5

Find the zeros of 8x+2+5x^2=4

The zeros of 8x+2+5x^2=4 are x=(-4+sqrt(26))/5 ,x=-(4+sqrt(26))/5

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8 x + 2 + 5 x^{2} = 4

x = \frac{- 4 + 26}{5}, x = - \frac{4 + 26}{5}

Decimal Notation

x = 0.21980 \dots, x = - 1.81980 \dots

Solution steps

8 x + 2 + 5 x^{2} = 4

Move

4

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

8^{2} - 4 \cdot 5 (- 2) = 226

x_{1, 2} = \frac{- 8 \pm 2 26}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 8 + 2 26}{2 \cdot 5}, x_{2} = \frac{- 8 - 2 26}{2 \cdot 5}

x = \frac{- 8 + 2 26}{2 \cdot 5} : \frac{- 4 + 26}{5}

x = \frac{- 8 - 2 26}{2 \cdot 5} : - \frac{4 + 26}{5}

The solutions to the quadratic equation are:

x = \frac{- 4 + 26}{5}, x = - \frac{4 + 26}{5}

2 x (9 - 5 x) = - 5 x (2 - 3 x)

(3 y - 5) (2 y + 7) = 0

\frac{x ^{2}}{2} + x - 4 = 0

2 n^{2} - 9 n - 2 = 0

25 t^{2} - 20 t + 4 = 0

What are the solutions to the equation 8x+2+5x^2=4 ?
The solutions to the equation 8x+2+5x^2=4 are x=(-4+sqrt(26))/5 ,x=-(4+sqrt(26))/5
Find the zeros of 8x+2+5x^2=4
The zeros of 8x+2+5x^2=4 are x=(-4+sqrt(26))/5 ,x=-(4+sqrt(26))/5