What are the solutions to the equation (2x^2+1)/5+(x+3)/(10)=1 ?

The solutions to the equation (2x^2+1)/5+(x+3)/(10)=1 are x=1,x=-5/4

Find the zeros of (2x^2+1)/5+(x+3)/(10)=1

The zeros of (2x^2+1)/5+(x+3)/(10)=1 are x=1,x=-5/4

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\frac{2 x ^{2} + 1}{5} + \frac{x + 3}{10} = 1

x = 1, x = - \frac{5}{4}

Decimal Notation

x = 1, x = - 1.25

Solution steps

\frac{2 x ^{2} + 1}{5} + \frac{x + 3}{10} = 1

Find Least Common Multiplier of

5, 10 : 10

Multiply by LCM=

10

\frac{2 x ^{2} + 1}{5} \cdot 10 + \frac{x + 3}{10} \cdot 10 = 1 \cdot 10

Simplify

2 (2 x^{2} + 1) + x + 3 = 10

Expand

2 (2 x^{2} + 1) + x + 3 : 4 x^{2} + x + 5

4 x^{2} + x + 5 = 10

Move

10

to the left side

4 x^{2} + x - 5 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 5) = 9

x_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 9}{2 \cdot 4}, x_{2} = \frac{- 1 - 9}{2 \cdot 4}

x = \frac{- 1 + 9}{2 \cdot 4} : 1

x = \frac{- 1 - 9}{2 \cdot 4} : - \frac{5}{4}

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{4}

x^{2} - 13 x - 14 = 0

6 x^{2} + 2 x + 4 = 0

x^{2} + 14 x - 70 = 9

- 9 x^{2} = - 864

\frac{x ^{2}}{9} + \frac{0 ^{2}}{16} = 1

What are the solutions to the equation (2x^2+1)/5+(x+3)/(10)=1 ?
The solutions to the equation (2x^2+1)/5+(x+3)/(10)=1 are x=1,x=-5/4
Find the zeros of (2x^2+1)/5+(x+3)/(10)=1
The zeros of (2x^2+1)/5+(x+3)/(10)=1 are x=1,x=-5/4