What are the solutions to the equation 2x^2-3x+10=7 ?

The solutions to the equation 2x^2-3x+10=7 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4

Find the zeros of 2x^2-3x+10=7

The zeros of 2x^2-3x+10=7 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4

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2 x^{2} - 3 x + 10 = 7

x = \frac{3}{4} + i \frac{15}{4}, x = \frac{3}{4} - i \frac{15}{4}

Solution steps

2 x^{2} - 3 x + 10 = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 15 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 15 i}{2 \cdot 2} : \frac{3}{4} + i \frac{15}{4}

x = \frac{- ( - 3 ) - 15 i}{2 \cdot 2} : \frac{3}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{15}{4}, x = \frac{3}{4} - i \frac{15}{4}

\frac{8 - 3 x}{2} = 5

1.34 d^{2} - 1.1 d = - 1.02

b^{3}

\frac{a - \frac{2 a - 1}{a}}{\frac{1 - a}{3 a}}

y^{2} + 8 y - 9 = 0

What are the solutions to the equation 2x^2-3x+10=7 ?
The solutions to the equation 2x^2-3x+10=7 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4
Find the zeros of 2x^2-3x+10=7
The zeros of 2x^2-3x+10=7 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4