What are the solutions to the equation 28+5t=(5+t)^2 ?

The solutions to the equation 28+5t=(5+t)^2 are t=(-5+sqrt(37))/2 ,t=(-5-sqrt(37))/2

Find the zeros of 28+5t=(5+t)^2

The zeros of 28+5t=(5+t)^2 are t=(-5+sqrt(37))/2 ,t=(-5-sqrt(37))/2

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28 + 5 t = (5 + t)^{2}

t = \frac{- 5 + 37}{2}, t = \frac{- 5 - 37}{2}

Decimal Notation

t = 0.54138 \dots, t = - 5.54138 \dots

Solution steps

28 + 5 t = (5 + t)^{2}

Expand

(5 + t)^{2} : 25 + 10 t + t^{2}

28 + 5 t = 25 + 10 t + t^{2}

Switch sides

25 + 10 t + t^{2} = 28 + 5 t

Move

5 t

to the left side

t^{2} + 5 t + 25 = 28

Move

28

to the left side

t^{2} + 5 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 3) = 37

t_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 5 + 37}{2 \cdot 1}, t_{2} = \frac{- 5 - 37}{2 \cdot 1}

t = \frac{- 5 + 37}{2 \cdot 1} : \frac{- 5 + 37}{2}

t = \frac{- 5 - 37}{2 \cdot 1} : \frac{- 5 - 37}{2}

The solutions to the quadratic equation are:

t = \frac{- 5 + 37}{2}, t = \frac{- 5 - 37}{2}

2 x^{2} = - 8 x - 7

(x - 2)^{2} = 0

\frac{1 + \frac{3}{3}}{1 - \frac{3}{3}}

23 - 4

15 x - 26 = 23 x - 10

What are the solutions to the equation 28+5t=(5+t)^2 ?
The solutions to the equation 28+5t=(5+t)^2 are t=(-5+sqrt(37))/2 ,t=(-5-sqrt(37))/2
Find the zeros of 28+5t=(5+t)^2
The zeros of 28+5t=(5+t)^2 are t=(-5+sqrt(37))/2 ,t=(-5-sqrt(37))/2