What are the solutions to the equation x(1+x)= 3/4 ?

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

Find the zeros of x(1+x)= 3/4

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

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x (1 + x) = \frac{3}{4}

x = \frac{1}{2}, x = - \frac{3}{2}

Decimal Notation

x = 0.5, x = - 1.5

Solution steps

x (1 + x) = \frac{3}{4}

Multiply both sides by

4

4 x (1 + x) = 3

Expand

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

4 x + 4 x^{2} = 3

Move

3

to the left side

4 x + 4 x^{2} - 3 = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} + 4 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 3) = 8

x_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 4 + 8}{2 \cdot 4}, x_{2} = \frac{- 4 - 8}{2 \cdot 4}

x = \frac{- 4 + 8}{2 \cdot 4} : \frac{1}{2}

x = \frac{- 4 - 8}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{3}{2}

(1 - a) + 3 (1 - a)^{2} = 0

26 \cdot 33

x^{2} + 5 x - 24 = 0

\frac{11}{15} = w - \frac{8}{15}

x^{2} + 10 x + 1 \geq 12

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