What are the solutions to the equation (((2x-1)^2))/3-((x^2-5))/8 = 7/2 ?

The solutions to the equation (((2x-1)^2))/3-((x^2-5))/8 = 7/2 are x= 61/29 ,x=-1

Find the zeros of (((2x-1)^2))/3-((x^2-5))/8 = 7/2

The zeros of (((2x-1)^2))/3-((x^2-5))/8 = 7/2 are x= 61/29 ,x=-1

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\frac{( ( 2 x - 1 ) ^{2} )}{3} - \frac{( x ^{2} - 5 )}{8} = \frac{7}{2}

x = \frac{61}{29}, x = - 1

Decimal Notation

x = 2.10344 \dots, x = - 1

Solution steps

\frac{( ( 2 x - 1 ) ^{2} )}{3} - \frac{( x ^{2} - 5 )}{8} = \frac{7}{2}

Find Least Common Multiplier of

3, 8, 2 : 24

Multiply by LCM=

24

\frac{( 2 x - 1 ) ^{2}}{3} \cdot 24 - \frac{x ^{2} - 5}{8} \cdot 24 = \frac{7}{2} \cdot 24

Simplify

8 (2 x - 1)^{2} - 3 (x^{2} - 5) = 84

Expand

8 (2 x - 1)^{2} - 3 (x^{2} - 5) : 29 x^{2} - 32 x + 23

29 x^{2} - 32 x + 23 = 84

Move

84

to the left side

29 x^{2} - 32 x - 61 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 29 ( - 61 )}{2 \cdot 29}

(- 32)^{2} - 4 \cdot 29 (- 61) = 90

x_{1, 2} = \frac{- ( - 32 ) \pm 90}{2 \cdot 29}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 90}{2 \cdot 29}, x_{2} = \frac{- ( - 32 ) - 90}{2 \cdot 29}

x = \frac{- ( - 32 ) + 90}{2 \cdot 29} : \frac{61}{29}

x = \frac{- ( - 32 ) - 90}{2 \cdot 29} : - 1

The solutions to the quadratic equation are:

x = \frac{61}{29}, x = - 1

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

\frac{x ^{2}}{3} = - x^{2} + 4 x

x^{2} + 2 x - 60 = 0

2.4 + t + 0.1 t^{2} = 0

5 (x^{2} - 1) = 4 (x^{2} + 1)

What are the solutions to the equation (((2x-1)^2))/3-((x^2-5))/8 = 7/2 ?
The solutions to the equation (((2x-1)^2))/3-((x^2-5))/8 = 7/2 are x= 61/29 ,x=-1
Find the zeros of (((2x-1)^2))/3-((x^2-5))/8 = 7/2
The zeros of (((2x-1)^2))/3-((x^2-5))/8 = 7/2 are x= 61/29 ,x=-1