What are the solutions to the equation 1%=(r^2)/4+r ?

The solutions to the equation 1%=(r^2)/4+r are r=2(sqrt(1%+1)-1),r=-2(sqrt(1%+1)+1)

Find the zeros of 1%=(r^2)/4+r

The zeros of 1%=(r^2)/4+r are r=2(sqrt(1%+1)-1),r=-2(sqrt(1%+1)+1)

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1% = \frac{r ^{2}}{4} + r

r = 2 (1% + 1 - 1), r = - 2 (1% + 1 + 1)

Solution steps

1% = \frac{r ^{2}}{4} + r

Multiply both sides by

4

4 \cdot 1% = r^{2} + 4 r

Switch sides

r^{2} + 4 r = 4 \cdot 1%

Move

4 \cdot 1%

to the left side

r^{2} + 4 r - 4 \cdot 1% = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 4 \cdot 1% )}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot (- 4 \cdot 1%) : 4 1 + 1%

r_{1, 2} = \frac{- 4 \pm 4 1 + 1%}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 4 1 + 1%}{2 \cdot 1}, r_{2} = \frac{- 4 - 4 1 + 1%}{2 \cdot 1}

r = \frac{- 4 + 4 1 + 1%}{2 \cdot 1} : 2 (1% + 1 - 1)

r = \frac{- 4 - 4 1 + 1%}{2 \cdot 1} : - 2 (1% + 1 + 1)

The solutions to the quadratic equation are:

r = 2 (1% + 1 - 1), r = - 2 (1% + 1 + 1)

3 d t - 21 d + 35 - 5 t

5 x^{2} - 3 x - 5 = 0

\frac{y}{4} - \frac{y + 1}{5} = \frac{y + 2}{8}

16 x^{3} - 9 x^{2} - 7 x

What are the solutions to the equation 1%=(r^2)/4+r ?
The solutions to the equation 1%=(r^2)/4+r are r=2(sqrt(1%+1)-1),r=-2(sqrt(1%+1)+1)
Find the zeros of 1%=(r^2)/4+r
The zeros of 1%=(r^2)/4+r are r=2(sqrt(1%+1)-1),r=-2(sqrt(1%+1)+1)