What are the solutions to the equation 2.5(x+1)^2=6x^2 ?

The solutions to the equation 2.5(x+1)^2=6x^2 are x=-(-5+2sqrt(15))/7 ,x=(5+2sqrt(15))/7

Find the zeros of 2.5(x+1)^2=6x^2

The zeros of 2.5(x+1)^2=6x^2 are x=-(-5+2sqrt(15))/7 ,x=(5+2sqrt(15))/7

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2.5 (x + 1)^{2} = 6 x^{2}

x = - \frac{- 5 + 2 15}{7}, x = \frac{5 + 2 15}{7}

Decimal Notation

x = - 0.39228 \dots, x = 1.82085 \dots

Solution steps

2.5 (x + 1)^{2} = 6 x^{2}

Multiply both sides by

10

25 (x + 1)^{2} = 60 x^{2}

Expand

25 (x + 1)^{2} : 25 x^{2} + 50 x + 25

25 x^{2} + 50 x + 25 = 60 x^{2}

Move

60 x^{2}

to the left side

- 35 x^{2} + 50 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 35 ) \cdot 25}{2 ( - 35 )}

5 0^{2} - 4 (- 35) \cdot 25 = 2015

x_{1, 2} = \frac{- 50 \pm 20 15}{2 ( - 35 )}

Separate the solutions

x_{1} = \frac{- 50 + 20 15}{2 ( - 35 )}, x_{2} = \frac{- 50 - 20 15}{2 ( - 35 )}

x = \frac{- 50 + 20 15}{2 ( - 35 )} : - \frac{- 5 + 2 15}{7}

x = \frac{- 50 - 20 15}{2 ( - 35 )} : \frac{5 + 2 15}{7}

The solutions to the quadratic equation are:

x = - \frac{- 5 + 2 15}{7}, x = \frac{5 + 2 15}{7}

x^{2} - 18 - 3 x = 0

R^{2} + 7 R + 12 = 0

4 y^{2} - 9 y + 2 = 0

10 x = x^{2} - 11

216 t^{2} - 42.6 t + 80.84 = 9607

What are the solutions to the equation 2.5(x+1)^2=6x^2 ?
The solutions to the equation 2.5(x+1)^2=6x^2 are x=-(-5+2sqrt(15))/7 ,x=(5+2sqrt(15))/7
Find the zeros of 2.5(x+1)^2=6x^2
The zeros of 2.5(x+1)^2=6x^2 are x=-(-5+2sqrt(15))/7 ,x=(5+2sqrt(15))/7