What are the solutions to the equation 2x^2-20x=-49 ?

The solutions to the equation 2x^2-20x=-49 are x=(10+sqrt(2))/2 ,x=(10-sqrt(2))/2

Find the zeros of 2x^2-20x=-49

The zeros of 2x^2-20x=-49 are x=(10+sqrt(2))/2 ,x=(10-sqrt(2))/2

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2 x^{2} - 20 x = - 49

x = \frac{10 + 2}{2}, x = \frac{10 - 2}{2}

Decimal Notation

x = 5.70710 \dots, x = 4.29289 \dots

Solution steps

2 x^{2} - 20 x = - 49

Move

49

to the left side

2 x^{2} - 20 x + 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 49}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 49 = 22

x_{1, 2} = \frac{- ( - 20 ) \pm 2 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 2 2}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 2 2}{2 \cdot 2}

x = \frac{- ( - 20 ) + 2 2}{2 \cdot 2} : \frac{10 + 2}{2}

x = \frac{- ( - 20 ) - 2 2}{2 \cdot 2} : \frac{10 - 2}{2}

The solutions to the quadratic equation are:

x = \frac{10 + 2}{2}, x = \frac{10 - 2}{2}

y^{2} + 75 = - 28 y

x, f (x + h) = 3 - 5 x - 2 x^{2}

10 x^{2} + 11 x - 90 = 0

x, f (x + 2) = 3 x^{2} + 4 x - 2

x^{2} - 6 x - 39 = 0

What are the solutions to the equation 2x^2-20x=-49 ?
The solutions to the equation 2x^2-20x=-49 are x=(10+sqrt(2))/2 ,x=(10-sqrt(2))/2
Find the zeros of 2x^2-20x=-49
The zeros of 2x^2-20x=-49 are x=(10+sqrt(2))/2 ,x=(10-sqrt(2))/2