What are the solutions to the equation a^2-13a+42=0 ?

The solutions to the equation a^2-13a+42=0 are a=7,a=6

Find the zeros of a^2-13a+42=0

The zeros of a^2-13a+42=0 are a=7,a=6

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a^{2} - 13 a + 42 = 0

a = 7, a = 6

Solution steps

a^{2} - 13 a + 42 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 42 = 1

a_{1, 2} = \frac{- ( - 13 ) \pm 1}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 13 ) + 1}{2 \cdot 1}, a_{2} = \frac{- ( - 13 ) - 1}{2 \cdot 1}

a = \frac{- ( - 13 ) + 1}{2 \cdot 1} : 7

a = \frac{- ( - 13 ) - 1}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

a = 7, a = 6

\frac{π}{2 π}

\frac{e ^{2 i} - e ^{- 2 i}}{2 i}

r^{2} + 2 = 0

\frac{2}{3} x + 3 = \frac{1}{4} x - 7

10 (3 x^{2} - 13 x) = - 40

What are the solutions to the equation a^2-13a+42=0 ?
The solutions to the equation a^2-13a+42=0 are a=7,a=6
Find the zeros of a^2-13a+42=0
The zeros of a^2-13a+42=0 are a=7,a=6