What are the solutions to the equation 5x^2-10x-3x+1=-7x^2+11 ?

The solutions to the equation 5x^2-10x-3x+1=-7x^2+11 are x=(13+sqrt(649))/(24),x=(13-sqrt(649))/(24)

Find the zeros of 5x^2-10x-3x+1=-7x^2+11

The zeros of 5x^2-10x-3x+1=-7x^2+11 are x=(13+sqrt(649))/(24),x=(13-sqrt(649))/(24)

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5 x^{2} - 10 x - 3 x + 1 = - 7 x^{2} + 11

x = \frac{13 + 649}{24}, x = \frac{13 - 649}{24}

Decimal Notation

x = 1.60314 \dots, x = - 0.51981 \dots

Solution steps

5 x^{2} - 10 x - 3 x + 1 = - 7 x^{2} + 11

Refine

5 x^{2} - 13 x + 1 = - 7 x^{2} + 11

Move

11

to the left side

5 x^{2} - 13 x - 10 = - 7 x^{2}

Move

7 x^{2}

to the left side

12 x^{2} - 13 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 12 ( - 10 )}{2 \cdot 12}

(- 13)^{2} - 4 \cdot 12 (- 10) = 649

x_{1, 2} = \frac{- ( - 13 ) \pm 649}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 649}{2 \cdot 12}, x_{2} = \frac{- ( - 13 ) - 649}{2 \cdot 12}

x = \frac{- ( - 13 ) + 649}{2 \cdot 12} : \frac{13 + 649}{24}

x = \frac{- ( - 13 ) - 649}{2 \cdot 12} : \frac{13 - 649}{24}

The solutions to the quadratic equation are:

x = \frac{13 + 649}{24}, x = \frac{13 - 649}{24}

x, x^{2} y - 6 x y + 8 y = 1

4 (x - 1)^{2} = 28

(X - 5)^{2} - 9 = 0

(5 x - 3)^{2} = - 32

8 w^{2} - 14 w + 5 = 0

What are the solutions to the equation 5x^2-10x-3x+1=-7x^2+11 ?
The solutions to the equation 5x^2-10x-3x+1=-7x^2+11 are x=(13+sqrt(649))/(24),x=(13-sqrt(649))/(24)
Find the zeros of 5x^2-10x-3x+1=-7x^2+11
The zeros of 5x^2-10x-3x+1=-7x^2+11 are x=(13+sqrt(649))/(24),x=(13-sqrt(649))/(24)