What are the solutions to the equation 3.2x^2-7.1x+2.5=0 ?

The solutions to the equation 3.2x^2-7.1x+2.5=0 are x=(71+sqrt(1841))/(64),x=(71-sqrt(1841))/(64)

Find the zeros of 3.2x^2-7.1x+2.5=0

The zeros of 3.2x^2-7.1x+2.5=0 are x=(71+sqrt(1841))/(64),x=(71-sqrt(1841))/(64)

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3.2 x^{2} - 7.1 x + 2.5 = 0

x = \frac{71 + 1841}{64}, x = \frac{71 - 1841}{64}

Decimal Notation

x = 1.77979 \dots, x = 0.43895 \dots

Solution steps

3.2 x^{2} - 7.1 x + 2.5 = 0

Multiply both sides by

10

32 x^{2} - 71 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 71 ) \pm ( - 71 ) ^{2} - 4 \cdot 32 \cdot 25}{2 \cdot 32}

(- 71)^{2} - 4 \cdot 32 \cdot 25 = 1841

x_{1, 2} = \frac{- ( - 71 ) \pm 1841}{2 \cdot 32}

Separate the solutions

x_{1} = \frac{- ( - 71 ) + 1841}{2 \cdot 32}, x_{2} = \frac{- ( - 71 ) - 1841}{2 \cdot 32}

x = \frac{- ( - 71 ) + 1841}{2 \cdot 32} : \frac{71 + 1841}{64}

x = \frac{- ( - 71 ) - 1841}{2 \cdot 32} : \frac{71 - 1841}{64}

The solutions to the quadratic equation are:

x = \frac{71 + 1841}{64}, x = \frac{71 - 1841}{64}

- 12 x^{2} + 48 x - 16 = 0

4 (x - 1)^{2} = (2 - x)^{2}

2 x^{2} + 19 x + 45 = 0

t^{2} - 2 t - 24 = 0

x^{2} - (k - 3) x + k = 0

What are the solutions to the equation 3.2x^2-7.1x+2.5=0 ?
The solutions to the equation 3.2x^2-7.1x+2.5=0 are x=(71+sqrt(1841))/(64),x=(71-sqrt(1841))/(64)
Find the zeros of 3.2x^2-7.1x+2.5=0
The zeros of 3.2x^2-7.1x+2.5=0 are x=(71+sqrt(1841))/(64),x=(71-sqrt(1841))/(64)