What is -x^2+5x+3>= 0 ?

The solution to -x^2+5x+3>= 0 is (-sqrt(37)+5)/2 <= x<= (sqrt(37)+5)/2

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-x^2+5x+3>= 0

- x^{2} + 5 x + 3 \geq 0

\frac{- 37 + 5}{2} \leq x \leq \frac{37 + 5}{2}

Interval Notation

[\frac{- 37 + 5}{2}, \frac{37 + 5}{2}]

Decimal Notation

- 0.54138 \dots \leq x \leq 5.54138 \dots

Solution steps

- x^{2} + 5 x + 3 \geq 0

Complete the square

- x^{2} + 5 x + 3 : - (x - \frac{5}{2})^{2} + \frac{37}{4}

- (x - \frac{5}{2})^{2} + \frac{37}{4} \geq 0

Move

\frac{37}{4}

to the right side

- (x - \frac{5}{2})^{2} \geq - \frac{37}{4}

Multiply both sides by

- 1

(x - \frac{5}{2})^{2} \leq \frac{37}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{37}{4} \leq x - \frac{5}{2} \leq \frac{37}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{37}{4} \leq x - \frac{5}{2} and x - \frac{5}{2} \leq \frac{37}{4}

- \frac{37}{4} \leq x - \frac{5}{2} : x \geq \frac{- 37 + 5}{2}

x - \frac{5}{2} \leq \frac{37}{4} : x \leq \frac{37 + 5}{2}

Combine the intervals

x \geq \frac{- 37 + 5}{2} and x \leq \frac{37 + 5}{2}

Merge Overlapping Intervals

\frac{- 37 + 5}{2} \leq x \leq \frac{37 + 5}{2}

\frac{m}{m ^{3}}

\frac{2}{9} - (- \frac{1}{15})

\frac{( - 1 - ( - 1 ) )}{0.001 - 0}

(x^{3})^{2}

t^{2} + 4 t - 25 = 0

What is -x^2+5x+3>= 0 ?
The solution to -x^2+5x+3>= 0 is (-sqrt(37)+5)/2 <= x<= (sqrt(37)+5)/2