What are the solutions to the equation 2x^2-3x+15=0 ?

The solutions to the equation 2x^2-3x+15=0 are x= 3/4+i(sqrt(111))/4 ,x= 3/4-i(sqrt(111))/4

Find the zeros of 2x^2-3x+15=0

The zeros of 2x^2-3x+15=0 are x= 3/4+i(sqrt(111))/4 ,x= 3/4-i(sqrt(111))/4

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2 x^{2} - 3 x + 15 = 0

x = \frac{3}{4} + i \frac{111}{4}, x = \frac{3}{4} - i \frac{111}{4}

Solution steps

2 x^{2} - 3 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 15}{2 \cdot 2}

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 15 : 111 i

x_{1, 2} = \frac{- ( - 3 ) \pm 111 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 111 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 111 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 111 i}{2 \cdot 2} : \frac{3}{4} + i \frac{111}{4}

x = \frac{- ( - 3 ) - 111 i}{2 \cdot 2} : \frac{3}{4} - i \frac{111}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{111}{4}, x = \frac{3}{4} - i \frac{111}{4}

\frac{1 0 ^{- 9}}{2}

∣ x - 2 ∣ > 2

x^{2} + x = 6

2 x^{3} - 98 x

3 x - 4 = 11

What are the solutions to the equation 2x^2-3x+15=0 ?
The solutions to the equation 2x^2-3x+15=0 are x= 3/4+i(sqrt(111))/4 ,x= 3/4-i(sqrt(111))/4
Find the zeros of 2x^2-3x+15=0
The zeros of 2x^2-3x+15=0 are x= 3/4+i(sqrt(111))/4 ,x= 3/4-i(sqrt(111))/4