What are the solutions to the equation x(5+12x)=28 ?

The solutions to the equation x(5+12x)=28 are x= 4/3 ,x=-7/4

Find the zeros of x(5+12x)=28

The zeros of x(5+12x)=28 are x= 4/3 ,x=-7/4

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x (5 + 12 x) = 28

x = \frac{4}{3}, x = - \frac{7}{4}

Decimal Notation

x = 1.33333 \dots, x = - 1.75

Solution steps

x (5 + 12 x) = 28

Expand

x (5 + 12 x) : 5 x + 12 x^{2}

5 x + 12 x^{2} = 28

Move

28

to the left side

5 x + 12 x^{2} - 28 = 0

Write in the standard form

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

12 x^{2} + 5 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 12 ( - 28 )}{2 \cdot 12}

5^{2} - 4 \cdot 12 (- 28) = 37

x_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 5 + 37}{2 \cdot 12}, x_{2} = \frac{- 5 - 37}{2 \cdot 12}

x = \frac{- 5 + 37}{2 \cdot 12} : \frac{4}{3}

x = \frac{- 5 - 37}{2 \cdot 12} : - \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{4}{3}, x = - \frac{7}{4}

f (a) = \frac{6 a ^{2} - 7 a + 2}{3 a - 2}

Y = - 6 + (1)

\frac{y}{2} + 22 = 38

\frac{k ( k + 1 )}{2} + k + 1

4 (5 x + 3) = 14 x + 30

What are the solutions to the equation x(5+12x)=28 ?
The solutions to the equation x(5+12x)=28 are x= 4/3 ,x=-7/4
Find the zeros of x(5+12x)=28
The zeros of x(5+12x)=28 are x= 4/3 ,x=-7/4