What are the solutions to the equation 4z^2-5z+3=0 ?

The solutions to the equation 4z^2-5z+3=0 are z= 5/8+i(sqrt(23))/8 ,z= 5/8-i(sqrt(23))/8

Find the zeros of 4z^2-5z+3=0

The zeros of 4z^2-5z+3=0 are z= 5/8+i(sqrt(23))/8 ,z= 5/8-i(sqrt(23))/8

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4 z^{2} - 5 z + 3 = 0

z = \frac{5}{8} + i \frac{23}{8}, z = \frac{5}{8} - i \frac{23}{8}

Solution steps

4 z^{2} - 5 z + 3 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 3 : 23 i

z_{1, 2} = \frac{- ( - 5 ) \pm 23 i}{2 \cdot 4}

Separate the solutions

z_{1} = \frac{- ( - 5 ) + 23 i}{2 \cdot 4}, z_{2} = \frac{- ( - 5 ) - 23 i}{2 \cdot 4}

z = \frac{- ( - 5 ) + 23 i}{2 \cdot 4} : \frac{5}{8} + i \frac{23}{8}

z = \frac{- ( - 5 ) - 23 i}{2 \cdot 4} : \frac{5}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

z = \frac{5}{8} + i \frac{23}{8}, z = \frac{5}{8} - i \frac{23}{8}

x, x^{2} + 2 x y - y^{2} = 0

x + 1 = (x + 1)^{2}

3 t^{2} - 2 t - 1 = 0

x^{2} - 13 x + 56 = 0

(2 x - 3) (x + 5) = 7

What are the solutions to the equation 4z^2-5z+3=0 ?
The solutions to the equation 4z^2-5z+3=0 are z= 5/8+i(sqrt(23))/8 ,z= 5/8-i(sqrt(23))/8
Find the zeros of 4z^2-5z+3=0
The zeros of 4z^2-5z+3=0 are z= 5/8+i(sqrt(23))/8 ,z= 5/8-i(sqrt(23))/8