What is (2x-5)^2>= 72+(-4+x)^2 ?

The solution to (2x-5)^2>= 72+(-4+x)^2 is x = 7

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(2x-5)^2>= 72+(-4+x)^2

(2 x - 5)^{2} \geq 72 + (- 4 + x)^{2}

x \leq - 3 or x \geq 7

Interval Notation

(- \infty, - 3] \cup [7, \infty)

Solution steps

(2 x - 5)^{2} \geq 72 + (- 4 + x)^{2}

Rewrite in standard form

x^{2} - 4 x - 21 \geq 0

Factor

x^{2} - 4 x - 21 : (x + 3) (x - 7)

(x + 3) (x - 7) \geq 0

Identify the intervals

x \leq - 3 or x \geq 7

\frac{16}{9}

\frac{c ^{2}}{20} - \frac{c}{4} + \frac{1}{5} = 0

t^{2} - t + 2 = 0

3 x^{2} - 5 x - 8 = 0

2000 (1 + \frac{0.09}{12})^{60}

What is (2x-5)^2>= 72+(-4+x)^2 ?
The solution to (2x-5)^2>= 72+(-4+x)^2 is x<=-3\lor x>= 7