What are the solutions to the equation 2x^2-3x+13=0 ?

The solutions to the equation 2x^2-3x+13=0 are x= 3/4+i(sqrt(95))/4 ,x= 3/4-i(sqrt(95))/4

Find the zeros of 2x^2-3x+13=0

The zeros of 2x^2-3x+13=0 are x= 3/4+i(sqrt(95))/4 ,x= 3/4-i(sqrt(95))/4

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2 x^{2} - 3 x + 13 = 0

x = \frac{3}{4} + i \frac{95}{4}, x = \frac{3}{4} - i \frac{95}{4}

Solution steps

2 x^{2} - 3 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 13}{2 \cdot 2}

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 13 : 95 i

x_{1, 2} = \frac{- ( - 3 ) \pm 95 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 95 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 95 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 95 i}{2 \cdot 2} : \frac{3}{4} + i \frac{95}{4}

x = \frac{- ( - 3 ) - 95 i}{2 \cdot 2} : \frac{3}{4} - i \frac{95}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{95}{4}, x = \frac{3}{4} - i \frac{95}{4}

b^{2} - 2 b + 1

4 x^{2} - 7 x + 10 = 0

7 = 5 - 2 x

\frac{i ^{5} - i ^{- 8}}{2 i}

- (x - 1)^{5} (x + 1)^{3} \leq 0

What are the solutions to the equation 2x^2-3x+13=0 ?
The solutions to the equation 2x^2-3x+13=0 are x= 3/4+i(sqrt(95))/4 ,x= 3/4-i(sqrt(95))/4
Find the zeros of 2x^2-3x+13=0
The zeros of 2x^2-3x+13=0 are x= 3/4+i(sqrt(95))/4 ,x= 3/4-i(sqrt(95))/4