What are the solutions to the equation 4t(3-t)=4t+3 ?

The solutions to the equation 4t(3-t)=4t+3 are t= 1/2 ,t= 3/2

Find the zeros of 4t(3-t)=4t+3

The zeros of 4t(3-t)=4t+3 are t= 1/2 ,t= 3/2

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4 t (3 - t) = 4 t + 3

t = \frac{1}{2}, t = \frac{3}{2}

Decimal Notation

t = 0.5, t = 1.5

Solution steps

4 t (3 - t) = 4 t + 3

Expand

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

12 t - 4 t^{2} = 4 t + 3

Move

3

to the left side

12 t - 4 t^{2} - 3 = 4 t

Move

4 t

to the left side

- 4 t^{2} + 8 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 3) = 4

t_{1, 2} = \frac{- 8 \pm 4}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 8 + 4}{2 ( - 4 )}, t_{2} = \frac{- 8 - 4}{2 ( - 4 )}

t = \frac{- 8 + 4}{2 ( - 4 )} : \frac{1}{2}

t = \frac{- 8 - 4}{2 ( - 4 )} : \frac{3}{2}

The solutions to the quadratic equation are:

t = \frac{1}{2}, t = \frac{3}{2}

(a + c) (a - c) = 0

x, ((p - 3) (p + 5)) \cdot x^{2} - 4 x + 8 = 0

x lo g_{10} (10) (x - 3) = 2

x, 2 x^{2} - 4 x y + 2 y^{2} - 5 x - 5 = 0

x, z^{2} = \frac{324}{972} (x^{2} + y^{2})

What are the solutions to the equation 4t(3-t)=4t+3 ?
The solutions to the equation 4t(3-t)=4t+3 are t= 1/2 ,t= 3/2
Find the zeros of 4t(3-t)=4t+3
The zeros of 4t(3-t)=4t+3 are t= 1/2 ,t= 3/2