What are the solutions to the equation 5a^2-142a+425=0 ?

The solutions to the equation 5a^2-142a+425=0 are a=25,a= 17/5

Find the zeros of 5a^2-142a+425=0

The zeros of 5a^2-142a+425=0 are a=25,a= 17/5

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5 a^{2} - 142 a + 425 = 0

a = 25, a = \frac{17}{5}

Decimal Notation

a = 25, a = 3.4

Solution steps

5 a^{2} - 142 a + 425 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 142 ) \pm ( - 142 ) ^{2} - 4 \cdot 5 \cdot 425}{2 \cdot 5}

(- 142)^{2} - 4 \cdot 5 \cdot 425 = 108

a_{1, 2} = \frac{- ( - 142 ) \pm 108}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- ( - 142 ) + 108}{2 \cdot 5}, a_{2} = \frac{- ( - 142 ) - 108}{2 \cdot 5}

a = \frac{- ( - 142 ) + 108}{2 \cdot 5} : 25

a = \frac{- ( - 142 ) - 108}{2 \cdot 5} : \frac{17}{5}

The solutions to the quadratic equation are:

a = 25, a = \frac{17}{5}

x^{2} - 3 x - 40 = 0

((a - 1) x + 2 a - 1) x - 3 a x = 2 a - 3

(5 \frac{x}{3}) (5 \frac{x}{4}) = 5^{5}

0 = 3 t^{2} - 6 t + 12

x^{2} = 12 x - 35

What are the solutions to the equation 5a^2-142a+425=0 ?
The solutions to the equation 5a^2-142a+425=0 are a=25,a= 17/5
Find the zeros of 5a^2-142a+425=0
The zeros of 5a^2-142a+425=0 are a=25,a= 17/5