What are the solutions to the equation 4x^2-3x=-3 ?

The solutions to the equation 4x^2-3x=-3 are x= 3/8+i(sqrt(39))/8 ,x= 3/8-i(sqrt(39))/8

Find the zeros of 4x^2-3x=-3

The zeros of 4x^2-3x=-3 are x= 3/8+i(sqrt(39))/8 ,x= 3/8-i(sqrt(39))/8

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4 x^{2} - 3 x = - 3

x = \frac{3}{8} + i \frac{39}{8}, x = \frac{3}{8} - i \frac{39}{8}

Solution steps

4 x^{2} - 3 x = - 3

Move

3

to the left side

4 x^{2} - 3 x + 3 = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 4 \cdot 3 : 39 i

x_{1, 2} = \frac{- ( - 3 ) \pm 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 39 i}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 39 i}{2 \cdot 4}

x = \frac{- ( - 3 ) + 39 i}{2 \cdot 4} : \frac{3}{8} + i \frac{39}{8}

x = \frac{- ( - 3 ) - 39 i}{2 \cdot 4} : \frac{3}{8} - i \frac{39}{8}

The solutions to the quadratic equation are:

x = \frac{3}{8} + i \frac{39}{8}, x = \frac{3}{8} - i \frac{39}{8}

3 x^{2} - 4 x - 4 = 3

9 r^{2} + 6 r + 4 = 0

16 (4 x - 5)^{2} = 9

(\frac{2}{5})^{2} + x^{2} = 1

x, x^{2} - 6 x + 8 = 0

What are the solutions to the equation 4x^2-3x=-3 ?
The solutions to the equation 4x^2-3x=-3 are x= 3/8+i(sqrt(39))/8 ,x= 3/8-i(sqrt(39))/8
Find the zeros of 4x^2-3x=-3
The zeros of 4x^2-3x=-3 are x= 3/8+i(sqrt(39))/8 ,x= 3/8-i(sqrt(39))/8