What are the solutions to the equation e=x^2-3x+5 ?

The solutions to the equation e=x^2-3x+5 are x= 3/2+i(sqrt(11-4e))/2 ,x= 3/2-i(sqrt(11-4e))/2

Find the zeros of e=x^2-3x+5

The zeros of e=x^2-3x+5 are x= 3/2+i(sqrt(11-4e))/2 ,x= 3/2-i(sqrt(11-4e))/2

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e = x^{2} - 3 x + 5

x = \frac{3}{2} + i \frac{11 - 4 e}{2}, x = \frac{3}{2} - i \frac{11 - 4 e}{2}

Solution steps

e = x^{2} - 3 x + 5

Switch sides

x^{2} - 3 x + 5 = e

Move

e

to the left side

x^{2} - 3 x + 5 - e = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot (5 - e) : i - 4 e + 11

x_{1, 2} = \frac{- ( - 3 ) \pm i - 4 e + 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + i - 4 e + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - i - 4 e + 11}{2 \cdot 1}

x = \frac{- ( - 3 ) + i - 4 e + 11}{2 \cdot 1} : \frac{3}{2} + i \frac{11 - 4 e}{2}

x = \frac{- ( - 3 ) - i - 4 e + 11}{2 \cdot 1} : \frac{3}{2} - i \frac{11 - 4 e}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{11 - 4 e}{2}, x = \frac{3}{2} - i \frac{11 - 4 e}{2}

(10 - x) (\frac{x}{4 - 3}) = 2

\frac{x}{2} = \frac{( x ^{2} )}{7}

x, 2 x \cdot y^{1} + y = \frac{( 2 x ^{2} )}{( y ^{3} )}

\frac{( x ^{2} + 8 )}{2} = 3 x

6 x^{2} - 17 = 7

What are the solutions to the equation e=x^2-3x+5 ?
The solutions to the equation e=x^2-3x+5 are x= 3/2+i(sqrt(11-4e))/2 ,x= 3/2-i(sqrt(11-4e))/2
Find the zeros of e=x^2-3x+5
The zeros of e=x^2-3x+5 are x= 3/2+i(sqrt(11-4e))/2 ,x= 3/2-i(sqrt(11-4e))/2