What are the solutions to the equation 2x^2-18x=40 ?

The solutions to the equation 2x^2-18x=40 are x=(9+sqrt(161))/2 ,x=(9-sqrt(161))/2

Find the zeros of 2x^2-18x=40

The zeros of 2x^2-18x=40 are x=(9+sqrt(161))/2 ,x=(9-sqrt(161))/2

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2 x^{2} - 18 x = 40

x = \frac{9 + 161}{2}, x = \frac{9 - 161}{2}

Decimal Notation

x = 10.84428 \dots, x = - 1.84428 \dots

Solution steps

2 x^{2} - 18 x = 40

Move

40

to the left side

2 x^{2} - 18 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 2 ( - 40 )}{2 \cdot 2}

(- 18)^{2} - 4 \cdot 2 (- 40) = 2161

x_{1, 2} = \frac{- ( - 18 ) \pm 2 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 161}{2 \cdot 2}, x_{2} = \frac{- ( - 18 ) - 2 161}{2 \cdot 2}

x = \frac{- ( - 18 ) + 2 161}{2 \cdot 2} : \frac{9 + 161}{2}

x = \frac{- ( - 18 ) - 2 161}{2 \cdot 2} : \frac{9 - 161}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 161}{2}, x = \frac{9 - 161}{2}

(x - 7)^{2} + (x - 7) = 6

a x^{2} - b x - c = 0

- 3 x^{2} - 12 x = 4

2 \cdot 3 + 12 = x^{2} - 13

(2 x - 7) (x + 9) = 0

What are the solutions to the equation 2x^2-18x=40 ?
The solutions to the equation 2x^2-18x=40 are x=(9+sqrt(161))/2 ,x=(9-sqrt(161))/2
Find the zeros of 2x^2-18x=40
The zeros of 2x^2-18x=40 are x=(9+sqrt(161))/2 ,x=(9-sqrt(161))/2