What are the solutions to the equation (3x-5)^2+7(3x-5)+12=0 ?

The solutions to the equation (3x-5)^2+7(3x-5)+12=0 are x= 2/3 ,x= 1/3

Find the zeros of (3x-5)^2+7(3x-5)+12=0

The zeros of (3x-5)^2+7(3x-5)+12=0 are x= 2/3 ,x= 1/3

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(3 x - 5)^{2} + 7 (3 x - 5) + 12 = 0

x = \frac{2}{3}, x = \frac{1}{3}

Decimal Notation

x = 0.66666 \dots, x = 0.33333 \dots

Solution steps

(3 x - 5)^{2} + 7 (3 x - 5) + 12 = 0

Expand

(3 x - 5)^{2} + 7 (3 x - 5) + 12 : 9 x^{2} - 9 x + 2

9 x^{2} - 9 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 9 \cdot 2}{2 \cdot 9}

(- 9)^{2} - 4 \cdot 9 \cdot 2 = 3

x_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3}{2 \cdot 9}, x_{2} = \frac{- ( - 9 ) - 3}{2 \cdot 9}

x = \frac{- ( - 9 ) + 3}{2 \cdot 9} : \frac{2}{3}

x = \frac{- ( - 9 ) - 3}{2 \cdot 9} : \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = \frac{1}{3}

∣ 8 - 4 x ∣ \leq 3

\frac{- n ^{2} + 16 n - 63}{n ^{2} - 2 n - 35}

\frac{1}{3} \cdot 60

4 + 3 (10 n + 6) \leq n - 3 - 10

x^{2} - 18 x + 74 = 0

What are the solutions to the equation (3x-5)^2+7(3x-5)+12=0 ?
The solutions to the equation (3x-5)^2+7(3x-5)+12=0 are x= 2/3 ,x= 1/3
Find the zeros of (3x-5)^2+7(3x-5)+12=0
The zeros of (3x-5)^2+7(3x-5)+12=0 are x= 2/3 ,x= 1/3