How do you solve for x in simplest radical form: $2 {(x + 3)}^{2} + 10 = 66$ $2(x+3)^2+10=66$ ?

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Answer 1 · 2015-04-14T18:04:09

$2 {(x + 3)}^{2} + 10 = 66$ $2(x+3)^2+10=66$

${(x + 3)}^{2} + 5 = 33$ $(x+3)^2 + 5 = 33$

$x^{2} + 6 x + 9 + 5 - 33 = 0$ $x^2+6x+9+5-33=0$

$x^{2} + 6 x - 19 = 0$ $x^2+6x-19=0$

Using the quadratic root formula: $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2} a$ $(-b+-sqrt(b^2-4ac))/2a$

$x = \frac{- 6 \pm \sqrt{36 + 76}}{2}$ $x=(-6+-sqrt(36+76))/2$

$x = \frac{- 6 \pm 4 \sqrt{7}}{2}$ $x= (-6+-4sqrt(7))/2$

$x = - 3 - 2 \sqrt{7}$ $x=-3-2sqrt(7)$
or
$x = - 3 + 2 \sqrt{7}$ $x=-3+2sqrt(7)$

Answer 2 · 2015-04-14T22:06:12

Alternative solution:

$2 {(x + 3)}^{2} + 10 = 66$ $2(x+3)^2 +10 = 66$

$2 {(x + 3)}^{2} = 66 - 10$ $2(x+3)^2 = 66 -10$ $ss$ $color(white)"ss"$ added $- 10$ $-10$ on both sides

$2 {(x + 3)}^{2} = 56$ $2(x+3)^2 = 56$

${(x + 3)}^{2} = \frac{56}{2}$ $(x+3)^2 = 56/2$ $ssssssss$ $color(white)"ssssssss"$ multiplied $\frac{1}{2}$ $1/2$ on both sides

${(x + 3)}^{2} = 28$ $(x+3)^2 = 28$

$x + 3 = \pm \sqrt{28}$ $x+3 = +- sqrt 28$ $ssssss$ $color(white)"ssssss"$ there are 2 numbers whise square is 28

$x = - 3 \pm \sqrt{28}$ $x = -3 +- sqrt 28$ $ss$ $color(white)"ss"$ added $- 3$ $-3$ on both sides

$x = - 3 \pm \sqrt{4 \cdot 7} = - 3 \pm \sqrt{4} \sqrt{7} = - 3 \pm 2 \sqrt{7}$ $x = -3 +- sqrt(4*7) = -3 +- sqrt4 sqrt7 = -3 +- 2sqrt7$

How do you solve for x in simplest radical form: 2(x+3)^2+10=662(x+3)2+10=662(x+3)^2+10=66?