What are the solutions to the equation 0.8x^2+4.8x-1.6=1.2 ?

The solutions to the equation 0.8x^2+4.8x-1.6=1.2 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2

Find the zeros of 0.8x^2+4.8x-1.6=1.2

The zeros of 0.8x^2+4.8x-1.6=1.2 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2

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0.8 x^{2} + 4.8 x - 1.6 = 1.2

x = \frac{- 6 + 5 2}{2}, x = - \frac{6 + 5 2}{2}

Decimal Notation

x = 0.53553 \dots, x = - 6.53553 \dots

Solution steps

0.8 x^{2} + 4.8 x - 1.6 = 1.2

Multiply both sides by

10

8 x^{2} + 48 x - 16 = 12

Move

12

to the left side

8 x^{2} + 48 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 8 ( - 28 )}{2 \cdot 8}

4 8^{2} - 4 \cdot 8 (- 28) = 402

x_{1, 2} = \frac{- 48 \pm 40 2}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 48 + 40 2}{2 \cdot 8}, x_{2} = \frac{- 48 - 40 2}{2 \cdot 8}

x = \frac{- 48 + 40 2}{2 \cdot 8} : \frac{- 6 + 5 2}{2}

x = \frac{- 48 - 40 2}{2 \cdot 8} : - \frac{6 + 5 2}{2}

The solutions to the quadratic equation are:

x = \frac{- 6 + 5 2}{2}, x = - \frac{6 + 5 2}{2}

6 b^{2} - 13 b + 6

(- 5 x^{5} y^{2})^{5}

5 \cdot 1 0^{4}

3 x^{2} - 5 x - 4 = 0

5^{- 2} = \frac{1}{25}

What are the solutions to the equation 0.8x^2+4.8x-1.6=1.2 ?
The solutions to the equation 0.8x^2+4.8x-1.6=1.2 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2
Find the zeros of 0.8x^2+4.8x-1.6=1.2
The zeros of 0.8x^2+4.8x-1.6=1.2 are x=(-6+5sqrt(2))/2 ,x=-(6+5sqrt(2))/2