What are the solutions to the equation -12w^2+20w-3=0 ?

The solutions to the equation -12w^2+20w-3=0 are w= 1/6 ,w= 3/2

Find the zeros of -12w^2+20w-3=0

The zeros of -12w^2+20w-3=0 are w= 1/6 ,w= 3/2

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- 12 w^{2} + 20 w - 3 = 0

w = \frac{1}{6}, w = \frac{3}{2}

Solution steps

- 12 w^{2} + 20 w - 3 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 12 ) ( - 3 )}{2 ( - 12 )}

2 0^{2} - 4 (- 12) (- 3) = 16

w_{1, 2} = \frac{- 20 \pm 16}{2 ( - 12 )}

Separate the solutions

w_{1} = \frac{- 20 + 16}{2 ( - 12 )}, w_{2} = \frac{- 20 - 16}{2 ( - 12 )}

w = \frac{- 20 + 16}{2 ( - 12 )} : \frac{1}{6}

w = \frac{- 20 - 16}{2 ( - 12 )} : \frac{3}{2}

The solutions to the quadratic equation are:

w = \frac{1}{6}, w = \frac{3}{2}

∣ 2 x + 3 ∣ \leq 9

\frac{5}{3} - \frac{7}{4} x = \frac{1}{4}

952 = 16 t^{2}

2 t > 8

3 x^{2} + 13 x = 10

What are the solutions to the equation -12w^2+20w-3=0 ?
The solutions to the equation -12w^2+20w-3=0 are w= 1/6 ,w= 3/2
Find the zeros of -12w^2+20w-3=0
The zeros of -12w^2+20w-3=0 are w= 1/6 ,w= 3/2