What are the solutions to the equation 0.32x^2-1.2x+1=0 ?

The solutions to the equation 0.32x^2-1.2x+1=0 are x= 5/2 ,x= 5/4

Find the zeros of 0.32x^2-1.2x+1=0

The zeros of 0.32x^2-1.2x+1=0 are x= 5/2 ,x= 5/4

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0.32 x^{2} - 1.2 x + 1 = 0

x = \frac{5}{2}, x = \frac{5}{4}

Decimal Notation

x = 2.5, x = 1.25

Solution steps

0.32 x^{2} - 1.2 x + 1 = 0

Multiply both sides by

100

32 x^{2} - 120 x + 100 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 \cdot 32 \cdot 100}{2 \cdot 32}

(- 120)^{2} - 4 \cdot 32 \cdot 100 = 40

x_{1, 2} = \frac{- ( - 120 ) \pm 40}{2 \cdot 32}

Separate the solutions

x_{1} = \frac{- ( - 120 ) + 40}{2 \cdot 32}, x_{2} = \frac{- ( - 120 ) - 40}{2 \cdot 32}

x = \frac{- ( - 120 ) + 40}{2 \cdot 32} : \frac{5}{2}

x = \frac{- ( - 120 ) - 40}{2 \cdot 32} : \frac{5}{4}

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = \frac{5}{4}

a x^{2} + (a + b) x + b = 0

x + 3 = 39 - x^{2}

y = y (4 - y)

x^{2} - 10 x = - 22

a = (x - 6) (x - 20) - (x - 3) 2 + 50

What are the solutions to the equation 0.32x^2-1.2x+1=0 ?
The solutions to the equation 0.32x^2-1.2x+1=0 are x= 5/2 ,x= 5/4
Find the zeros of 0.32x^2-1.2x+1=0
The zeros of 0.32x^2-1.2x+1=0 are x= 5/2 ,x= 5/4