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Popular >

solvefor t,t=(2a(bt^2+n))/3+d/2

Solution

solve for

t, t = \frac{2 a ( b t ^{2} + n )}{3} + \frac{d}{2}

Solution

t = \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}, t = \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}; 4 ab \neq = 0

Solution steps

t = \frac{2 a ( b t ^{2} + n )}{3} + \frac{d}{2}

Find Least Common Multiplier of

3, 2 : 6

Multiply by LCM=

6

t \cdot 6 = \frac{2 a ( b t ^{2} + n )}{3} \cdot 6 + \frac{d}{2} \cdot 6

Simplify

6 t = 4 a (n + t^{2} b) + 3 d

Expand

4 a (n + t^{2} b) + 3 d : 4 na + 4 t^{2} ab + 3 d

6 t = 4 na + 4 t^{2} ab + 3 d

Switch sides

4 na + 4 t^{2} ab + 3 d = 6 t

Move

6 t

to the left side

4 na + 4 t^{2} ab + 3 d - 6 t = 0

Write in the standard form

a x^{2} + b x + c = 0

4 ab t^{2} - 6 t + 4 na + 3 d = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ab ( 4 na + 3 d )}{2 \cdot 4 ab}

Simplify

(- 6)^{2} - 4 \cdot 4 ab (4 na + 3 d) : 2 9 - 4 ab (3 d + 4 na)

t_{1, 2} = \frac{- ( - 6 ) \pm 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}; 4 ab \neq = 0

Separate the solutions

t_{1} = \frac{- ( - 6 ) + 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}, t_{2} = \frac{- ( - 6 ) - 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}

t = \frac{- ( - 6 ) + 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab} : \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}

t = \frac{- ( - 6 ) - 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab} : \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}

The solutions to the quadratic equation are:

t = \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}, t = \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}; 4 ab \neq = 0

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