What are the solutions to the equation a^2+14a-51=0 ?

The solutions to the equation a^2+14a-51=0 are a=3,a=-17

Find the zeros of a^2+14a-51=0

The zeros of a^2+14a-51=0 are a=3,a=-17

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a^{2} + 14 a - 51 = 0

a = 3, a = - 17

Solution steps

a^{2} + 14 a - 51 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot ( - 51 )}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot (- 51) = 20

a_{1, 2} = \frac{- 14 \pm 20}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 14 + 20}{2 \cdot 1}, a_{2} = \frac{- 14 - 20}{2 \cdot 1}

a = \frac{- 14 + 20}{2 \cdot 1} : 3

a = \frac{- 14 - 20}{2 \cdot 1} : - 17

The solutions to the quadratic equation are:

a = 3, a = - 17

(\frac{x}{4})^{2}

(- \frac{1}{4})^{- 3}

21 6^{\frac{1}{3}}

4.5 c - 2.8 = 3 (1.5 c - 1.9)

n^{2} - 18 n + 66 = 0

What are the solutions to the equation a^2+14a-51=0 ?
The solutions to the equation a^2+14a-51=0 are a=3,a=-17
Find the zeros of a^2+14a-51=0
The zeros of a^2+14a-51=0 are a=3,a=-17