What are the solutions to the equation t^2=28-3t ?

The solutions to the equation t^2=28-3t are t=4,t=-7

Find the zeros of t^2=28-3t

The zeros of t^2=28-3t are t=4,t=-7

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t^{2} = 28 - 3 t

t = 4, t = - 7

Solution steps

t^{2} = 28 - 3 t

Move

3 t

to the left side

t^{2} + 3 t = 28

Move

28

to the left side

t^{2} + 3 t - 28 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

t_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 11}{2 \cdot 1}, t_{2} = \frac{- 3 - 11}{2 \cdot 1}

t = \frac{- 3 + 11}{2 \cdot 1} : 4

t = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

t = 4, t = - 7

7 x^{2} - 9 x + 2

(9 x - 1)^{2} = 1

(9 x + 59) + (6 x + 16) = 180

(0.1 70)^{2} + 1.69

x^{5} + x^{4} - 13 x^{3} - 13 x^{2} + 36 x + 36

What are the solutions to the equation t^2=28-3t ?
The solutions to the equation t^2=28-3t are t=4,t=-7
Find the zeros of t^2=28-3t
The zeros of t^2=28-3t are t=4,t=-7