What are the solutions to the equation (x^2-3x+4)=0 ?

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

Find the zeros of (x^2-3x+4)=0

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

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

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

Solution steps

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

Refine

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} + 18 x + 36 = 0

λ^{2} + 6 λ + 5 = 0

x^{2} - 7 = - 14

x^{2} - 7 = - 12

s^{2} - 4 s - 4 = 0

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