What are the solutions to the equation a^2-3a-1=6 ?

The solutions to the equation a^2-3a-1=6 are a=(3+sqrt(37))/2 ,a=(3-sqrt(37))/2

Find the zeros of a^2-3a-1=6

The zeros of a^2-3a-1=6 are a=(3+sqrt(37))/2 ,a=(3-sqrt(37))/2

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a^{2} - 3 a - 1 = 6

a = \frac{3 + 37}{2}, a = \frac{3 - 37}{2}

Decimal Notation

a = 4.54138 \dots, a = - 1.54138 \dots

Solution steps

a^{2} - 3 a - 1 = 6

Move

6

to the left side

a^{2} - 3 a - 7 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 7) = 37

a_{1, 2} = \frac{- ( - 3 ) \pm 37}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 3 ) + 37}{2 \cdot 1}, a_{2} = \frac{- ( - 3 ) - 37}{2 \cdot 1}

a = \frac{- ( - 3 ) + 37}{2 \cdot 1} : \frac{3 + 37}{2}

a = \frac{- ( - 3 ) - 37}{2 \cdot 1} : \frac{3 - 37}{2}

The solutions to the quadratic equation are:

a = \frac{3 + 37}{2}, a = \frac{3 - 37}{2}

x, z = x^{2} + 4 y^{2}

x, y = x^{2} - 3 x + 2

\frac{1}{2} x^{2} - 5 = 5

r, S = 2 π r h + 2 π r^{2}

A (x) = - x (x - 80)

What are the solutions to the equation a^2-3a-1=6 ?
The solutions to the equation a^2-3a-1=6 are a=(3+sqrt(37))/2 ,a=(3-sqrt(37))/2
Find the zeros of a^2-3a-1=6
The zeros of a^2-3a-1=6 are a=(3+sqrt(37))/2 ,a=(3-sqrt(37))/2