What are the solutions to the equation 25x-19=3(x^2+3) ?

The solutions to the equation 25x-19=3(x^2+3) are x=7,x= 4/3

Find the zeros of 25x-19=3(x^2+3)

The zeros of 25x-19=3(x^2+3) are x=7,x= 4/3

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25 x - 19 = 3 (x^{2} + 3)

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

Decimal Notation

x = 7, x = 1.33333 \dots

Solution steps

25 x - 19 = 3 (x^{2} + 3)

Expand

3 (x^{2} + 3) : 3 x^{2} + 9

25 x - 19 = 3 x^{2} + 9

Switch sides

3 x^{2} + 9 = 25 x - 19

Move

19

to the left side

3 x^{2} + 28 = 25 x

Move

25 x

to the left side

3 x^{2} + 28 - 25 x = 0

Write in the standard form

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

3 x^{2} - 25 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 25 ) \pm ( - 25 ) ^{2} - 4 \cdot 3 \cdot 28}{2 \cdot 3}

(- 25)^{2} - 4 \cdot 3 \cdot 28 = 17

x_{1, 2} = \frac{- ( - 25 ) \pm 17}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 25 ) + 17}{2 \cdot 3}, x_{2} = \frac{- ( - 25 ) - 17}{2 \cdot 3}

x = \frac{- ( - 25 ) + 17}{2 \cdot 3} : 7

x = \frac{- ( - 25 ) - 17}{2 \cdot 3} : \frac{4}{3}

The solutions to the quadratic equation are:

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

x^{2} + x - 12 = 0

(2 x - 5)^{2} = 169

x^{3} - 3 x - 2 = (x - a)^{2} (x + b)

x^{2} - 1 = 0, y^{2} + z = 0, z^{2} + y = 0

(x + 3) (x - 2) = 4 (x - 2)

What are the solutions to the equation 25x-19=3(x^2+3) ?
The solutions to the equation 25x-19=3(x^2+3) are x=7,x= 4/3
Find the zeros of 25x-19=3(x^2+3)
The zeros of 25x-19=3(x^2+3) are x=7,x= 4/3