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

The solutions to the equation y^2-3y+4=0 are y= 3/2+i(sqrt(7))/2 ,y= 3/2-i(sqrt(7))/2

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

The zeros of y^2-3y+4=0 are y= 3/2+i(sqrt(7))/2 ,y= 3/2-i(sqrt(7))/2

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

y = \frac{3}{2} + i \frac{7}{2}, y = \frac{3}{2} - i \frac{7}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 4 : 7 i

y_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 1}

y = \frac{- ( - 3 ) + 7 i}{2 \cdot 1} : \frac{3}{2} + i \frac{7}{2}

y = \frac{- ( - 3 ) - 7 i}{2 \cdot 1} : \frac{3}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

y = \frac{3}{2} + i \frac{7}{2}, y = \frac{3}{2} - i \frac{7}{2}

6 x^{\frac{3}{2}} - 18 x^{\frac{1}{2}} + 12 x^{- \frac{1}{2}}

2 (\frac{20}{29}) (\frac{21}{29})

∣ x ∣ + 8 > 9

\frac{4 - ( - 5 )}{4 - ( - 6 )}

8 x^{2} - 6 x = 0

What are the solutions to the equation y^2-3y+4=0 ?
The solutions to the equation y^2-3y+4=0 are y= 3/2+i(sqrt(7))/2 ,y= 3/2-i(sqrt(7))/2
Find the zeros of y^2-3y+4=0
The zeros of y^2-3y+4=0 are y= 3/2+i(sqrt(7))/2 ,y= 3/2-i(sqrt(7))/2