What are the solutions to the equation 3c^2-18c+24=((-2)-(2))/((4)-(2)) ?

The solutions to the equation 3c^2-18c+24=((-2)-(2))/((4)-(2)) are c=(9+sqrt(3))/3 ,c=(9-sqrt(3))/3

Find the zeros of 3c^2-18c+24=((-2)-(2))/((4)-(2))

The zeros of 3c^2-18c+24=((-2)-(2))/((4)-(2)) are c=(9+sqrt(3))/3 ,c=(9-sqrt(3))/3

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3 c^{2} - 18 c + 24 = \frac{( - 2 ) - ( 2 )}{( 4 ) - ( 2 )}

c = \frac{9 + 3}{3}, c = \frac{9 - 3}{3}

Decimal Notation

c = 3.57735 \dots, c = 2.42264 \dots

Solution steps

3 c^{2} - 18 c + 24 = \frac{( - 2 ) - ( 2 )}{( 4 ) - ( 2 )}

Simplify

(4) - (2) : 2

3 c^{2} - 18 c + 24 = \frac{( - 2 ) - ( 2 )}{2}

Multiply both sides by

2

6 c^{2} - 36 c + 48 = - 4

Move

4

to the left side

6 c^{2} - 36 c + 52 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 6 \cdot 52}{2 \cdot 6}

(- 36)^{2} - 4 \cdot 6 \cdot 52 = 43

c_{1, 2} = \frac{- ( - 36 ) \pm 4 3}{2 \cdot 6}

Separate the solutions

c_{1} = \frac{- ( - 36 ) + 4 3}{2 \cdot 6}, c_{2} = \frac{- ( - 36 ) - 4 3}{2 \cdot 6}

c = \frac{- ( - 36 ) + 4 3}{2 \cdot 6} : \frac{9 + 3}{3}

c = \frac{- ( - 36 ) - 4 3}{2 \cdot 6} : \frac{9 - 3}{3}

The solutions to the quadratic equation are:

c = \frac{9 + 3}{3}, c = \frac{9 - 3}{3}

x^{2} + x - 34 = 0

m^{2} + 6 m + 1 = 0

t^{2} + 8 t + 1 = 0

- 0.25 x^{2} + 4 x - 9 = x - 2.71

x^{2} - 2 x + 81 = 0

What are the solutions to the equation 3c^2-18c+24=((-2)-(2))/((4)-(2)) ?
The solutions to the equation 3c^2-18c+24=((-2)-(2))/((4)-(2)) are c=(9+sqrt(3))/3 ,c=(9-sqrt(3))/3
Find the zeros of 3c^2-18c+24=((-2)-(2))/((4)-(2))
The zeros of 3c^2-18c+24=((-2)-(2))/((4)-(2)) are c=(9+sqrt(3))/3 ,c=(9-sqrt(3))/3