What are the solutions to the equation 4w(w-3)=24 ?

The solutions to the equation 4w(w-3)=24 are w=(3+sqrt(33))/2 ,w=(3-sqrt(33))/2

Find the zeros of 4w(w-3)=24

The zeros of 4w(w-3)=24 are w=(3+sqrt(33))/2 ,w=(3-sqrt(33))/2

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4 w (w - 3) = 24

w = \frac{3 + 33}{2}, w = \frac{3 - 33}{2}

Decimal Notation

w = 4.37228 \dots, w = - 1.37228 \dots

Solution steps

4 w (w - 3) = 24

Expand

4 w (w - 3) : 4 w^{2} - 12 w

4 w^{2} - 12 w = 24

Move

24

to the left side

4 w^{2} - 12 w - 24 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 ( - 24 )}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 (- 24) = 433

w_{1, 2} = \frac{- ( - 12 ) \pm 4 33}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- ( - 12 ) + 4 33}{2 \cdot 4}, w_{2} = \frac{- ( - 12 ) - 4 33}{2 \cdot 4}

w = \frac{- ( - 12 ) + 4 33}{2 \cdot 4} : \frac{3 + 33}{2}

w = \frac{- ( - 12 ) - 4 33}{2 \cdot 4} : \frac{3 - 33}{2}

The solutions to the quadratic equation are:

w = \frac{3 + 33}{2}, w = \frac{3 - 33}{2}

\frac{b ^{\frac{2}{3}}}{b ^{\frac{1}{7}}}

- (2 x - 3) \leq 4 - (2 x + 4)

8 + t < 10

w^{2} + w - 6 = 0

3 + k \geq 1 and \frac{k}{8} < 1

What are the solutions to the equation 4w(w-3)=24 ?
The solutions to the equation 4w(w-3)=24 are w=(3+sqrt(33))/2 ,w=(3-sqrt(33))/2
Find the zeros of 4w(w-3)=24
The zeros of 4w(w-3)=24 are w=(3+sqrt(33))/2 ,w=(3-sqrt(33))/2