Question #f107b

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Question #f107b

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Answer 1 · 2017-11-17T17:08:07

See explanation

Explanation:

$y^{2} - 14 y + 46 = 0$ $y^2-14y+46=0$

Lets identify a, b, and c

$a y^{2} + b y + c = 0$ $ay^2+by+c=0$
$a = 1$ $a=1$
$b = - 14$ $b=-14$
$c = 46$ $c=46$

Plug this into the quadratic formula-
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

$\frac{- (- 14) \pm \sqrt{{(- 14)}^{2} - 4 (1 \cdot 46)}}{2 (1)}$ $(-(-14)+-sqrt((-14)^2-4(1*46)))/(2(1))$
Now solve
$\frac{14 \pm \sqrt{196 - 184}}{2}$ $(14+-sqrt(196-184))/(2)$

$\frac{14 \pm \sqrt{12}}{2}$ $(14+-sqrt(12))/(2)$

$\frac{14 \pm 2 \sqrt{3}}{2}$ $(14+-2sqrt(3))/(2)$

y = 8.73205
y = 5.26794
or
y = $7 + \sqrt{3}$ $7+sqrt(3)$
y= $7 - \sqrt{3}$ $7-sqrt(3)$

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Question #f107b

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