What are the solutions to the equation T+3 T/2+((T+2))/2 T=4T^2 ?

The solutions to the equation T+3 T/2+((T+2))/2 T=4T^2 are T=0,T=1

Find the zeros of T+3 T/2+((T+2))/2 T=4T^2

The zeros of T+3 T/2+((T+2))/2 T=4T^2 are T=0,T=1

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T + 3 \frac{T}{2} + \frac{( T + 2 )}{2} T = 4 T^{2}

T = 0, T = 1

Solution steps

T + 3 \cdot \frac{T}{2} + \frac{( T + 2 )}{2} T = 4 T^{2}

Multiply both sides by

2

2 T + 3 T + T (T + 2) = 8 T^{2}

5 T + T (T + 2) = 8 T^{2}

Expand

5 T + T (T + 2) : T^{2} + 7 T

T^{2} + 7 T = 8 T^{2}

Move

8 T^{2}

to the left side

- 7 T^{2} + 7 T = 0

Solve with the quadratic formula

T_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 7 ) \cdot 0}{2 ( - 7 )}

7^{2} - 4 (- 7) \cdot 0 = 7

T_{1, 2} = \frac{- 7 \pm 7}{2 ( - 7 )}

Separate the solutions

T_{1} = \frac{- 7 + 7}{2 ( - 7 )}, T_{2} = \frac{- 7 - 7}{2 ( - 7 )}

T = \frac{- 7 + 7}{2 ( - 7 )} : 0

T = \frac{- 7 - 7}{2 ( - 7 )} : 1

The solutions to the quadratic equation are:

T = 0, T = 1

2 x - 3 (x + 1) \geq 3 x

(n^{4})^{\frac{3}{2}}

- 4 - 4 (2 p + 3) = - 3 (4 p - 4)

\frac{x - 1}{x + 1} \leq 0

x^{2} + 4 x - 9 = 5 x + 3

What are the solutions to the equation T+3 T/2+((T+2))/2 T=4T^2 ?
The solutions to the equation T+3 T/2+((T+2))/2 T=4T^2 are T=0,T=1
Find the zeros of T+3 T/2+((T+2))/2 T=4T^2
The zeros of T+3 T/2+((T+2))/2 T=4T^2 are T=0,T=1