What are the solutions to the equation a^2-17a+42=0 ?

The solutions to the equation a^2-17a+42=0 are a=14,a=3

Find the zeros of a^2-17a+42=0

The zeros of a^2-17a+42=0 are a=14,a=3

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a^{2} - 17 a + 42 = 0

a = 14, a = 3

Solution steps

a^{2} - 17 a + 42 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 17)^{2} - 4 \cdot 1 \cdot 42 = 11

a_{1, 2} = \frac{- ( - 17 ) \pm 11}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 17 ) + 11}{2 \cdot 1}, a_{2} = \frac{- ( - 17 ) - 11}{2 \cdot 1}

a = \frac{- ( - 17 ) + 11}{2 \cdot 1} : 14

a = \frac{- ( - 17 ) - 11}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

a = 14, a = 3

7 x^{2} - 7 x + 9 = 0

\frac{π}{4} + \frac{π}{3}

2 x^{2} + 8 x + 16

\frac{5 x - 9}{- 3} = - 7

What are the solutions to the equation a^2-17a+42=0 ?
The solutions to the equation a^2-17a+42=0 are a=14,a=3
Find the zeros of a^2-17a+42=0
The zeros of a^2-17a+42=0 are a=14,a=3