What are the solutions to the equation (3a-2)(2a+1)-a(a+12)=0 ?

The solutions to the equation (3a-2)(2a+1)-a(a+12)=0 are a=(13+sqrt(209))/(10),a=(13-sqrt(209))/(10)

Find the zeros of (3a-2)(2a+1)-a(a+12)=0

The zeros of (3a-2)(2a+1)-a(a+12)=0 are a=(13+sqrt(209))/(10),a=(13-sqrt(209))/(10)

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(3 a - 2) (2 a + 1) - a (a + 12) = 0

a = \frac{13 + 209}{10}, a = \frac{13 - 209}{10}

Decimal Notation

a = 2.74568 \dots, a = - 0.14568 \dots

Solution steps

(3 a - 2) (2 a + 1) - a (a + 12) = 0

Expand

(3 a - 2) (2 a + 1) - a (a + 12) : 5 a^{2} - 13 a - 2

5 a^{2} - 13 a - 2 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

(- 13)^{2} - 4 \cdot 5 (- 2) = 209

a_{1, 2} = \frac{- ( - 13 ) \pm 209}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- ( - 13 ) + 209}{2 \cdot 5}, a_{2} = \frac{- ( - 13 ) - 209}{2 \cdot 5}

a = \frac{- ( - 13 ) + 209}{2 \cdot 5} : \frac{13 + 209}{10}

a = \frac{- ( - 13 ) - 209}{2 \cdot 5} : \frac{13 - 209}{10}

The solutions to the quadratic equation are:

a = \frac{13 + 209}{10}, a = \frac{13 - 209}{10}

1 6^{x} + 2 0^{x} = 2 5^{x}

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

x^{2} = 2^{x}

2^{x} = 5^{x + 2}

x^{x} = 2

What are the solutions to the equation (3a-2)(2a+1)-a(a+12)=0 ?
The solutions to the equation (3a-2)(2a+1)-a(a+12)=0 are a=(13+sqrt(209))/(10),a=(13-sqrt(209))/(10)
Find the zeros of (3a-2)(2a+1)-a(a+12)=0
The zeros of (3a-2)(2a+1)-a(a+12)=0 are a=(13+sqrt(209))/(10),a=(13-sqrt(209))/(10)