What are the solutions to the equation 4x^2+33x=27 ?

The solutions to the equation 4x^2+33x=27 are x= 3/4 ,x=-9

Find the zeros of 4x^2+33x=27

The zeros of 4x^2+33x=27 are x= 3/4 ,x=-9

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4 x^{2} + 33 x = 27

x = \frac{3}{4}, x = - 9

Decimal Notation

x = 0.75, x = - 9

Solution steps

4 x^{2} + 33 x = 27

Move

27

to the left side

4 x^{2} + 33 x - 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 33 \pm 3 3 ^{2} - 4 \cdot 4 ( - 27 )}{2 \cdot 4}

3 3^{2} - 4 \cdot 4 (- 27) = 39

x_{1, 2} = \frac{- 33 \pm 39}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 33 + 39}{2 \cdot 4}, x_{2} = \frac{- 33 - 39}{2 \cdot 4}

x = \frac{- 33 + 39}{2 \cdot 4} : \frac{3}{4}

x = \frac{- 33 - 39}{2 \cdot 4} : - 9

The solutions to the quadratic equation are:

x = \frac{3}{4}, x = - 9

6 (x - 1) < 3 x - 5

21 = \frac{t ^{2} - t}{2}

(13 n^{2} + 11 n - 2 n^{4}) + (- 13 n^{2} - 3 n - 6 n^{4})

\frac{1}{9 ^{8}}

(x - 3 i) (x + 3 i)

What are the solutions to the equation 4x^2+33x=27 ?
The solutions to the equation 4x^2+33x=27 are x= 3/4 ,x=-9
Find the zeros of 4x^2+33x=27
The zeros of 4x^2+33x=27 are x= 3/4 ,x=-9