What are the solutions to the equation 36x^2+39x=12 ?

The solutions to the equation 36x^2+39x=12 are x= 1/4 ,x=-4/3

Find the zeros of 36x^2+39x=12

The zeros of 36x^2+39x=12 are x= 1/4 ,x=-4/3

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36 x^{2} + 39 x = 12

x = \frac{1}{4}, x = - \frac{4}{3}

Decimal Notation

x = 0.25, x = - 1.33333 \dots

Solution steps

36 x^{2} + 39 x = 12

Move

12

to the left side

36 x^{2} + 39 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 39 \pm 3 9 ^{2} - 4 \cdot 36 ( - 12 )}{2 \cdot 36}

3 9^{2} - 4 \cdot 36 (- 12) = 57

x_{1, 2} = \frac{- 39 \pm 57}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- 39 + 57}{2 \cdot 36}, x_{2} = \frac{- 39 - 57}{2 \cdot 36}

x = \frac{- 39 + 57}{2 \cdot 36} : \frac{1}{4}

x = \frac{- 39 - 57}{2 \cdot 36} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = - \frac{4}{3}

x^{2} - 6 + 4 = 0

0 = (x - 2) (2 x + 3)

\frac{2}{3} x^{2} - \frac{3}{2} x = 0

2 x^{2} + 16.38 x + 33.01 x = 0

x^{2} = 6 x + 5

What are the solutions to the equation 36x^2+39x=12 ?
The solutions to the equation 36x^2+39x=12 are x= 1/4 ,x=-4/3
Find the zeros of 36x^2+39x=12
The zeros of 36x^2+39x=12 are x= 1/4 ,x=-4/3