What are the solutions to the equation 12x^2+12=25x ?

The solutions to the equation 12x^2+12=25x are x= 4/3 ,x= 3/4

Find the zeros of 12x^2+12=25x

The zeros of 12x^2+12=25x are x= 4/3 ,x= 3/4

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12 x^{2} + 12 = 25 x

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

Decimal Notation

x = 1.33333 \dots, x = 0.75

Solution steps

12 x^{2} + 12 = 25 x

Move

25 x

to the left side

12 x^{2} + 12 - 25 x = 0

Write in the standard form

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

12 x^{2} - 25 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 25 ) \pm ( - 25 ) ^{2} - 4 \cdot 12 \cdot 12}{2 \cdot 12}

(- 25)^{2} - 4 \cdot 12 \cdot 12 = 7

x_{1, 2} = \frac{- ( - 25 ) \pm 7}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 25 ) + 7}{2 \cdot 12}, x_{2} = \frac{- ( - 25 ) - 7}{2 \cdot 12}

x = \frac{- ( - 25 ) + 7}{2 \cdot 12} : \frac{4}{3}

x = \frac{- ( - 25 ) - 7}{2 \cdot 12} : \frac{3}{4}

The solutions to the quadratic equation are:

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

n (t) = - 2.82 t^{2} + 25.74 t + 60.87

(x - 6) (x - 8) = 7 (x - 1) (x + 8) + x^{2}

x, 2 x^{2} + 3 y^{2} + z^{2} - 12 x + 12 y + 2 z + 28 = 0

x, (x - y)^{2} = 2 a y

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

What are the solutions to the equation 12x^2+12=25x ?
The solutions to the equation 12x^2+12=25x are x= 4/3 ,x= 3/4
Find the zeros of 12x^2+12=25x
The zeros of 12x^2+12=25x are x= 4/3 ,x= 3/4