What is -5y^2+8y+8>= 8y+3 ?

The solution to -5y^2+8y+8>= 8y+3 is -1<= y<= 1

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-5y^2+8y+8>= 8y+3

- 5 y^{2} + 8 y + 8 \geq 8 y + 3

- 1 \leq y \leq 1

Interval Notation

[- 1, 1]

Solution steps

- 5 y^{2} + 8 y + 8 \geq 8 y + 3

Rewrite in standard form

- y^{2} + 1 \geq 0

Move

1

to the right side

- y^{2} \geq - 1

Multiply both sides by

- 1

y^{2} \leq 1

For

u^{n} \leq a

, if

n

is even then

- 1 \leq y \leq 1

\frac{\frac{x ^{2}}{y ^{2}} - 1}{\frac{x}{y} + 1}

3 x^{2} + 13 x - 10 = 0

(x - 5)^{2} = - 36

\frac{2}{3} x - \frac{5}{4} = - \frac{5}{6} x + \frac{7}{6}

6 x \geq 2 x - 8