What are the solutions to the equation 12x^2-36x+27=0 ?

The solutions to the equation 12x^2-36x+27=0 are x= 3/2

Find the zeros of 12x^2-36x+27=0

The zeros of 12x^2-36x+27=0 are x= 3/2

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12 x^{2} - 36 x + 27 = 0

x = \frac{3}{2}

Decimal Notation

x = 1.5

Solution steps

12 x^{2} - 36 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 12 \cdot 27}{2 \cdot 12}

(- 36)^{2} - 4 \cdot 12 \cdot 27 = 0

x_{1, 2} = \frac{- ( - 36 ) \pm 0}{2 \cdot 12}

x = \frac{- ( - 36 )}{2 \cdot 12}

\frac{- ( - 36 )}{2 \cdot 12} = \frac{3}{2}

x = \frac{3}{2}

The solution to the quadratic equation is:

x = \frac{3}{2}

3 (2 x - 1) + 7 = 6 x - 3 - 2 (2 - 4 x)

x \leq 1

1 - r = - 5

y^{2} + 5 y - 36

\frac{2}{3} - 2

What are the solutions to the equation 12x^2-36x+27=0 ?
The solutions to the equation 12x^2-36x+27=0 are x= 3/2
Find the zeros of 12x^2-36x+27=0
The zeros of 12x^2-36x+27=0 are x= 3/2