How do you solve $10 {(x - 7)}^{2} = 440$ $10(x-7)^2=440$ ?

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Answer 1 · 2018-03-02T02:53:39

$x = 7 - 2 \sqrt{11}, 7 + 2 \sqrt{11}$ $x = 7-2sqrt(11), 7+2sqrt(11)$

Divide both sides by 10:

${(x - 7)}^{2} = 44$ $(x-7)^2=44$

Take the square root of both sides

$x - 7 = \pm \sqrt{44}$ $x-7=+-sqrt(44)$

Add seven to both sides

$x = 7 \pm \sqrt{44}$ $x=7+-sqrt(44)$

You can simplify the square root. A perfect square of $4$ $4$ goes in there, so

$x = 7 \pm 2 \sqrt{11}$ $x=7+-2sqrt(11)$

Answer = $7 - 2 \sqrt{11}, 7 + 2 \sqrt{11}$ $7-2sqrt(11), 7+2sqrt(11)$

How do you solve 10(x−7)2=44010(x-7)^2=440?