How do you solve $y^{2} - 4 y - 21 = 0$ $y^2 - 4y - 21 = 0$ using the quadratic formula?

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How do you solve $y^{2} - 4 y - 21 = 0$ $y^2 - 4y - 21 = 0$ using the quadratic formula?

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Answer 1 · 2018-05-20T21:56:56

$x_{1} = 7$ $x_1=7$
$x_{2} = - 3$ $x_2=-3$

Explanation:

So basically the quadratic equation is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
Here $x = y, a = 1, b = - 4, c = - 21$ $x=y, a=1, b=-4, c=-21$
Plug that in and we get:
$x = \frac{- (- 4) \pm \sqrt{4^{2} - 4 \cdot 1 \cdot (- 21)}}{2 \cdot 1}$ $x=(-(-4)+-sqrt(4^2-4*1*(-21)))/(2*1)$

$x = \frac{4 \pm \sqrt{16 + 84}}{2}$ $x=(4+-sqrt(16+84))/2$

$x = \frac{4 \pm \sqrt{100}}{2}$ $x=(4+-sqrt100)/2$

$x = \frac{4 \pm 10}{2}$ $x=(4+-10)/2$

$x = 2 \pm 5$ $x=2+-5$

There are two possible values for $x$ $x$ .
$x_{1} = 7$ $x_1=7$
$x_{2} = - 3$ $x_2=-3$

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How do you solve $y^{2} - 4 y - 21 = 0$ $y^2 - 4y - 21 = 0$ using the quadratic formula?

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How do you solve $y^{2} - 4 y - 21 = 0$ $y^2 - 4y - 21 = 0$ using the quadratic formula?