What are the solutions to the equation 10x^2+5(-5x)^2+4x(-5x)=110 ?

The solutions to the equation 10x^2+5(-5x)^2+4x(-5x)=110 are x=sqrt(22/23),x=-sqrt(22/23)

Find the zeros of 10x^2+5(-5x)^2+4x(-5x)=110

The zeros of 10x^2+5(-5x)^2+4x(-5x)=110 are x=sqrt(22/23),x=-sqrt(22/23)

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10 x^{2} + 5 (- 5 x)^{2} + 4 x (- 5 x) = 110

x = \frac{22}{23}, x = - \frac{22}{23}

Decimal Notation

x = 0.97801 \dots, x = - 0.97801 \dots

Solution steps

10 x^{2} + 5 (- 5 x)^{2} + 4 x (- 5 x) = 110

Expand

10 x^{2} + 5 (- 5 x)^{2} + 4 x (- 5 x) : 115 x^{2}

115 x^{2} = 110

Move

110

to the left side

115 x^{2} - 110 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 115 ( - 110 )}{2 \cdot 115}

0^{2} - 4 \cdot 115 (- 110) = 10506

x_{1, 2} = \frac{- 0 \pm 10 506}{2 \cdot 115}

Separate the solutions

x_{1} = \frac{- 0 + 10 506}{2 \cdot 115}, x_{2} = \frac{- 0 - 10 506}{2 \cdot 115}

x = \frac{- 0 + 10 506}{2 \cdot 115} : \frac{22}{23}

x = \frac{- 0 - 10 506}{2 \cdot 115} : - \frac{22}{23}

The solutions to the quadratic equation are:

x = \frac{22}{23}, x = - \frac{22}{23}

4 x^{2} - 8 x = 0

x^{2} - 2 x + 1 = 49

4 x = 12 x^{2}

2 x^{2} - 7 x - 99 = 0

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

What are the solutions to the equation 10x^2+5(-5x)^2+4x(-5x)=110 ?
The solutions to the equation 10x^2+5(-5x)^2+4x(-5x)=110 are x=sqrt(22/23),x=-sqrt(22/23)
Find the zeros of 10x^2+5(-5x)^2+4x(-5x)=110
The zeros of 10x^2+5(-5x)^2+4x(-5x)=110 are x=sqrt(22/23),x=-sqrt(22/23)