What are the solutions to the equation 8x^2-4x=18 ?

The solutions to the equation 8x^2-4x=18 are x=(1+sqrt(37))/4 ,x=(1-sqrt(37))/4

Find the zeros of 8x^2-4x=18

The zeros of 8x^2-4x=18 are x=(1+sqrt(37))/4 ,x=(1-sqrt(37))/4

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8 x^{2} - 4 x = 18

x = \frac{1 + 37}{4}, x = \frac{1 - 37}{4}

Decimal Notation

x = 1.77069 \dots, x = - 1.27069 \dots

Solution steps

8 x^{2} - 4 x = 18

Move

18

to the left side

8 x^{2} - 4 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 8 ( - 18 )}{2 \cdot 8}

(- 4)^{2} - 4 \cdot 8 (- 18) = 437

x_{1, 2} = \frac{- ( - 4 ) \pm 4 37}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 37}{2 \cdot 8}, x_{2} = \frac{- ( - 4 ) - 4 37}{2 \cdot 8}

x = \frac{- ( - 4 ) + 4 37}{2 \cdot 8} : \frac{1 + 37}{4}

x = \frac{- ( - 4 ) - 4 37}{2 \cdot 8} : \frac{1 - 37}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 37}{4}, x = \frac{1 - 37}{4}

(21 + 2 i) + (13 + 8 i)

4 m^{4} - 324

f = 2 (0.75)^{3}

x - 12 = 5 x^{2}

x^{2} + 1 \geq 0

What are the solutions to the equation 8x^2-4x=18 ?
The solutions to the equation 8x^2-4x=18 are x=(1+sqrt(37))/4 ,x=(1-sqrt(37))/4
Find the zeros of 8x^2-4x=18
The zeros of 8x^2-4x=18 are x=(1+sqrt(37))/4 ,x=(1-sqrt(37))/4