What are the solutions to the equation 0.2x^2-2.2x-3.8=0 ?

The solutions to the equation 0.2x^2-2.2x-3.8=0 are x=(11+sqrt(197))/2 ,x=(11-sqrt(197))/2

Find the zeros of 0.2x^2-2.2x-3.8=0

The zeros of 0.2x^2-2.2x-3.8=0 are x=(11+sqrt(197))/2 ,x=(11-sqrt(197))/2

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0.2 x^{2} - 2.2 x - 3.8 = 0

x = \frac{11 + 197}{2}, x = \frac{11 - 197}{2}

Decimal Notation

x = 12.51783 \dots, x = - 1.51783 \dots

Solution steps

0.2 x^{2} - 2.2 x - 3.8 = 0

Multiply both sides by

10

2 x^{2} - 22 x - 38 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 2 ( - 38 )}{2 \cdot 2}

(- 22)^{2} - 4 \cdot 2 (- 38) = 2197

x_{1, 2} = \frac{- ( - 22 ) \pm 2 197}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 2 197}{2 \cdot 2}, x_{2} = \frac{- ( - 22 ) - 2 197}{2 \cdot 2}

x = \frac{- ( - 22 ) + 2 197}{2 \cdot 2} : \frac{11 + 197}{2}

x = \frac{- ( - 22 ) - 2 197}{2 \cdot 2} : \frac{11 - 197}{2}

The solutions to the quadratic equation are:

x = \frac{11 + 197}{2}, x = \frac{11 - 197}{2}

x^{2} + 6 = 70

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

2 x^{2} + 4 x + 7

t, x = v t + \frac{1}{2} a t^{2}

x^{2} = 7^{8}

What are the solutions to the equation 0.2x^2-2.2x-3.8=0 ?
The solutions to the equation 0.2x^2-2.2x-3.8=0 are x=(11+sqrt(197))/2 ,x=(11-sqrt(197))/2
Find the zeros of 0.2x^2-2.2x-3.8=0
The zeros of 0.2x^2-2.2x-3.8=0 are x=(11+sqrt(197))/2 ,x=(11-sqrt(197))/2