What are the solutions to the equation 18x^2-24x+87=0 ?

The solutions to the equation 18x^2-24x+87=0 are x= 2/3+i(sqrt(158))/6 ,x= 2/3-i(sqrt(158))/6

Find the zeros of 18x^2-24x+87=0

The zeros of 18x^2-24x+87=0 are x= 2/3+i(sqrt(158))/6 ,x= 2/3-i(sqrt(158))/6

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18 x^{2} - 24 x + 87 = 0

x = \frac{2}{3} + i \frac{158}{6}, x = \frac{2}{3} - i \frac{158}{6}

Solution steps

18 x^{2} - 24 x + 87 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 18 \cdot 87}{2 \cdot 18}

Simplify

(- 24)^{2} - 4 \cdot 18 \cdot 87 : 6158 i

x_{1, 2} = \frac{- ( - 24 ) \pm 6 158 i}{2 \cdot 18}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 6 158 i}{2 \cdot 18}, x_{2} = \frac{- ( - 24 ) - 6 158 i}{2 \cdot 18}

x = \frac{- ( - 24 ) + 6 158 i}{2 \cdot 18} : \frac{2}{3} + i \frac{158}{6}

x = \frac{- ( - 24 ) - 6 158 i}{2 \cdot 18} : \frac{2}{3} - i \frac{158}{6}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{158}{6}, x = \frac{2}{3} - i \frac{158}{6}

- 1 \leq \frac{3}{4} x + 2 < 8

- 9 \leq 2 x + 7 \leq 9

4.5 (x - 9) = 2.96

18 x^{2} - (9 x + 2) (2 x - 1) = 1

\frac{28 - 7 k ^{2}}{k ^{2} - 4}

What are the solutions to the equation 18x^2-24x+87=0 ?
The solutions to the equation 18x^2-24x+87=0 are x= 2/3+i(sqrt(158))/6 ,x= 2/3-i(sqrt(158))/6
Find the zeros of 18x^2-24x+87=0
The zeros of 18x^2-24x+87=0 are x= 2/3+i(sqrt(158))/6 ,x= 2/3-i(sqrt(158))/6