What are the solutions to the equation (2a+3)(3a-1)=24a ?

The solutions to the equation (2a+3)(3a-1)=24a are a=3,a=-1/6

Find the zeros of (2a+3)(3a-1)=24a

The zeros of (2a+3)(3a-1)=24a are a=3,a=-1/6

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(2 a + 3) (3 a - 1) = 24 a

a = 3, a = - \frac{1}{6}

Decimal Notation

a = 3, a = - 0.16666 \dots

Solution steps

(2 a + 3) (3 a - 1) = 24 a

Expand

(2 a + 3) (3 a - 1) : 6 a^{2} + 7 a - 3

6 a^{2} + 7 a - 3 = 24 a

Move

24 a

to the left side

6 a^{2} - 17 a - 3 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 17)^{2} - 4 \cdot 6 (- 3) = 19

a_{1, 2} = \frac{- ( - 17 ) \pm 19}{2 \cdot 6}

Separate the solutions

a_{1} = \frac{- ( - 17 ) + 19}{2 \cdot 6}, a_{2} = \frac{- ( - 17 ) - 19}{2 \cdot 6}

a = \frac{- ( - 17 ) + 19}{2 \cdot 6} : 3

a = \frac{- ( - 17 ) - 19}{2 \cdot 6} : - \frac{1}{6}

The solutions to the quadratic equation are:

a = 3, a = - \frac{1}{6}

2 n^{2} - 9 = n^{2}

2 k^{2} = 3 - 5 k

- 5 n^{2} + 9 n + 1 = - 6 n^{2}

0 = 210 + 205.06 x + 4.9 x^{2}

x, x^{2} + y^{2} = C y

What are the solutions to the equation (2a+3)(3a-1)=24a ?
The solutions to the equation (2a+3)(3a-1)=24a are a=3,a=-1/6
Find the zeros of (2a+3)(3a-1)=24a
The zeros of (2a+3)(3a-1)=24a are a=3,a=-1/6